## PHY2403H Quantum Field Theory. Lecture 5: Klein-Gordon equation, Hamilton’s equations, SHOs, momentum space representation, raising and lowering operators. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class. Some additional side notes, but otherwise barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Canonical quantization

Last time we introduced a Lagrangian density associated with the Klein-Gordon equation (with a quadratic potential coupling)
\label{eqn:qftLecture5:20}
L = \int d^3 x
\lr{
\inv{2} \lr{\partial_0 \phi}^2 – \inv{2} \lr{\spacegrad \phi}^2 – \frac{m^2}{2} \phi^2 – \frac{\lambda}{4} \phi^4
}.

This Lagrangian density was related to the action by
\label{eqn:qftLecture5:40}
S = \int dt L = \int dt d^3 x \LL,

with momentum canonically conjugate to the field $$\phi$$ defined as
\label{eqn:qftLecture5:60}
\Pi(\Bx, t) = \frac{\delta \LL}{\delta \phidot(\Bx, t) } = \PD{\phidot(\Bx, t)}{\LL}

The Hamiltonian defined as
\label{eqn:qftLecture5:80}
H = \int d^3 x \lr{ \Pi(\Bx, t) \phidot(\Bx, t) – \LL },

led to
\label{eqn:qftLecture5:680}
H
= \int d^3 x
\lr{ \inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4 }.

Like the Lagrangian density, we may introduce a Hamiltonian density $$\mathcal{H}$$ as
\label{eqn:qftLecture5:100}
H = \int d^3 x \mathcal{H}(\Bx, t).

For our Klein-Gordon system, this is
\label{eqn:qftLecture5:120}
\mathcal{H}(\Bx, t) =
\inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4.

### Canonical Commutation Relations (CCR)

:

We quantize the system by promoting our fields to Heisenberg-Picture (HP) operators, and imposing commutation relations
\label{eqn:qftLecture5:140}
\antisymmetric{\hat{\Pi}(\Bx, t)}{\hat{\phi}(\By, t)} = -i \delta^3 (\Bx – \By)

This is in analogy to
\label{eqn:qftLecture5:160}
\antisymmetric{\hat{p}_i}{\hat{q}_j} = -i \delta_{ij},

To choose a representation, we may map the $$\Psi$$ of QM $$\rightarrow$$ to a wave functional $$\Psi[\phi]$$
\label{eqn:qftLecture5:180}
\hat{\phi}(\By, t) \Psi[\phi] = \phi(\By, t) \Psi[\phi]

This is similar to the QM wave functions
\label{eqn:qftLecture5:200}
\begin{aligned}
\hat{q}_i \Psi(\setlr{q}) &= q_i \Psi(q) \\
\hat{p}_i \Psi(\setlr{q}) &= -i \PD{q_i}{} \Psi(p)
\end{aligned}

Our momentum operator is quantized by expressing it in terms of a variational derivative
\label{eqn:qftLecture5:220}
\hat{\Pi}(\Bx, t) = -i \frac{\delta}{\delta \phi(\Bx, t)}.

(Fixme: I’m not really sure exactly what is meant by using the variation derivative $$\delta$$ notation here), and to
quantize the Hamiltonian we just add hats, assuming that our fields are all now HP operators
\label{eqn:qftLecture5:240}
\hat{\mathcal{H}}(\Bx, t)
=
\inv{2} \hat{\Pi}^2 + (\spacegrad \hat{\phi})^2 + \inv{2} m^2 \hat{\phi}^2 + \frac{\lambda}{4} \hat{\phi}^4.

### QM SHO review

Recall the QM SHO had a Hamiltonian
\label{eqn:qftLecture5:260}
\hat{H} = \inv{2} \hat{p}^2 + \inv{2} \omega^2 \hat{q}^2,

where
\label{eqn:qftLecture5:280}
\antisymmetric{\hat{p}}{\hat{q}} = -i,

and that
HP time evolution operators $$O$$ satisfied
\label{eqn:qftLecture5:700}
\ddt{\hatO} = i \antisymmetric{\hatH}{\hatO}.

In particular
\label{eqn:qftLecture5:300}
\begin{aligned}
\ddt{\hat{p}}
&= i \antisymmetric{\hat{H}}{\hatp} \\
&= i \frac{\omega^2}{2} \antisymmetric{\hat{q}^2}{\hatp} \\
&= i \frac{\omega^2}{2} (2 i \hat{q}) \\
&= -i \omega^2 \hat{q},
\end{aligned}

and
\label{eqn:qftLecture5:320}
\begin{aligned}
\ddt{\hat{q}}
&= i \antisymmetric{\hat{H}}{\hat{q}} \\
&= i \inv{2} \antisymmetric{\hatp^2}{\hat{q}} \\
&= \frac{i}{2}(-2 i \hatp ) \\
&= \hatp.
\end{aligned}

Applying the time evolution operator twice, we find
\label{eqn:qftLecture5:340}
\frac{d^2}{dt^2}{\hat{q}}
= \ddt{\hat{p}}
= – \omega^2 \hat{q}.

We see that the Heisenberg operators obey the classical equations of motion.

Now we want to try this with the quantized QFT fields we’ve promoted to operators
\label{eqn:qftLecture5:360}
\begin{aligned}
\ddt{\hat{\Pi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\Pi}(\Bx, t)} \\
&=
i \int d^3 y \inv{2} \antisymmetric{ \lr{\spacegrad \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
+
i \int d^3 y \frac{m^2}{2} \antisymmetric{ \phihat(\By)^2 }{\hat{\Pi}(\Bx) }
+
i \frac{\lambda}{4} \int d^3 \antisymmetric{ \phihat(\By)^4 }{\hat{\Pi}(\Bx) }
\end{aligned}

Starting with the non-gradient commutators, and utilizing the HP field analogues of the relations $$\antisymmetric{\hat{q}^n}{\hatp} = n i \hat{q}^{n-1}$$, we find
\label{eqn:qftLecture5:780}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 2 i \phihat(\By) \delta^3(\Bx – \By)
= 2 i \phihat(\Bx).

\label{eqn:qftLecture5:740}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^4 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 4 i \phihat(\By)^3 \delta^3(\Bx – \By)
= 4 i \phihat(\Bx)^3.

For the gradient commutators, we have more work. Prof Poppitz blitzed through that, just calling it integration by parts. I had trouble seeing what he was doing, so here’s a more explicit dumb expansion required to calculate the commutator
\label{eqn:qftLecture5:720}
\begin{aligned}
\int d^3 y (\spacegrad \phihat(\By))^2 \hat{\Pi}(\Bx)
&=
\int d^3 y
&=
\int d^3 y
\lr{ \spacegrad (\phihat(\By) \hat{\Pi}(\Bx)) } \\
&=
\int d^3 y
\lr{ \spacegrad (\hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By)) } \\
&=
\int d^3 y
\Biglr{
+ i
} \\
&=
\int d^3 y
\Biglr{
\spacegrad \lr{ \hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By) } \cdot \spacegrad \phihat(\By)
+ i
} \\
&=
\int d^3 y
\hat{\Pi}(\Bx)
\lr{
}
+ 2 i
\int d^3 y
&=
\int d^3 y
+
2 i
\int d^3 y

2 i
\int d^3 y
\delta^3(\Bx – \By) \spacegrad^2 \phihat(\By) \\
&=
\int d^3 y
+
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

\end{aligned}

Here we take advantage of the fact that the derivative operators $$\spacegrad = \spacegrad_\By$$ commute with $$\hat{\Pi}(\Bx)$$, and use the identity
$$\spacegrad \cdot (a \spacegrad b) = (\spacegrad a) \cdot (\spacegrad b) + a \spacegrad^2 b$$, so the commutator is
\label{eqn:qftLecture5:800}
\begin{aligned}
&=
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

&=

\end{aligned}

where the boundary integral is presumed to be zero (without enough justification.) All the pieces can now be put back together
\label{eqn:qftLecture5:820}
\ddt{} \hat{\Pi}(\Bx, t)
=

m^2 \phihat(\Bx, t)

\lambda \phihat^3(\Bx, t).

Now, for the $$\phihat$$ time evolution, which is much easier
\label{eqn:qftLecture5:380}
\begin{aligned}
\ddt{\hat{\phi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\phi}(\Bx, t)} \\
&= i \inv{2} \int d^3 y \antisymmetric{\hat{\Pi}^2(\By)}{\hat{\phi}(\Bx)} \\
&= i \inv{2} \int d^3 y (-2 i) \hat{\Pi}(\By, t) \delta^3(\Bx – \By) \\
&= \hat{\Pi}(\Bx, t)
\end{aligned}

\label{eqn:qftLecture5:400}
\frac{d^2}{dt^2}{\hat{\phi}}(\Bx, t)
=
-m^2 \phi – \lambda \phihat^3.

That is
\label{eqn:qftLecture5:420}
\ddot{\phihat} – \spacegrad^2 \phihat + m^2 \phihat + \lambda \phihat^3 = 0,

which is the classical Euler-Lagrange equation, also obeyed by the Heisenberg operator $$\phi(\Bx, t)$$. When $$\lambda = 0$$ this is the Klein-Gordon equation.

## Momentum space representation.

Dropping hats, we now consider the momentum space representation of our operators, as determined by Fourier transform pairs
\label{eqn:qftLecture5:440}
\begin{aligned}
\phi(\Bx, t) &= \int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx} \tilde{\phi}(\Bp, t) \\
\tilde{\phi}(\Bp, t) &= \int d^3 x e^{-i \Bp \cdot \Bx} \phi(\Bx, t)
\end{aligned}

We can discover a representation of the delta function by applying these both in turn
\label{eqn:qftLecture5:480}
\tilde{\phi}(\Bp, t)
= \int d^3 x e^{-i \Bp \cdot \Bx} \int \frac{d^3 q}{(2 \pi)^3} e^{i \Bq \cdot \Bx} \tilde{\phi}(\Bq, t)

so
\label{eqn:qftLecture5:500}
\boxed{
\int d^3 x e^{i \BA \cdot \Bx} = (2 \pi)^3 \delta^3(\BA)
}

Also observe that $$\phi^\conj(\Bx, t) = \phi(\Bx, t)$$ iff $$\tilde{\phi}(\Bp, t) = \tilde{\phi}^\conj(-\Bp, t)$$.

We want the EOM for $$\tilde{\phi}(\Bp, t)$$ where the operator obeys the KG equation
\label{eqn:qftLecture5:520}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi(\Bx, t) = 0

Inserting the transform relation \ref{eqn:qftLecture5:440} we get
\label{eqn:qftLecture5:540}
\int \frac{d^3 p}{(2 \pi)^3} e^{i \Bp \cdot \Bx}
\lr{
\ddot{\tilde{\phi}}(\Bp, t) + \lr{ \Bp^2 + m^2 }
\tilde{\phi}(\Bp, t)
}
= 0,

or
\label{eqn:qftLecture5:580}
\boxed{
\ddot{\tilde{\phi}}(\Bp, t) = – \omega_\Bp^2 \,\tilde{\phi}(\Bp, t),
}

where
\label{eqn:qftLecture5:560}
\omega_\Bp = \sqrt{ \Bp^2 + m^2 }.

The Fourier components of the HP operators are SHOs!

As we have SHO’s and know how to deal with these in QM, we use the same strategy, introducing raising and lowering operators
\label{eqn:qftLecture5:600}
\tilde{\phi}(\Bp, t) = \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{-i \omega_\Bp t } a_\Bp + e^{i \omega_\Bp t} a^\dagger_{-\Bp}
}

Observe that
\label{eqn:qftLecture5:840}
\begin{aligned}
\tilde{\phi}^\dagger(-\Bp, t)
&= \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{i \omega_\Bp t } a^\dagger_{-\Bp} + e^{-i \omega_\Bp t} a_{\Bp} } \\
&=
\tilde{\phi}(\Bp, t),
\end{aligned}

or
\label{eqn:qftLecture5:620}
\tilde{\phi}^\dagger(\Bp, t) = \tilde{\phi}(-\Bp, t),

so $$\phi(\Bp, t)$$ has a real representation in terms of $$a_\Bp$$.

We will find (Wednesday) that
\label{eqn:qftLecture5:640}
\antisymmetric{a_\Bq}{a^+_\Bp} = \delta^3(\Bp – \Bq) (2 \pi)^3.

These are equivalent to
\label{eqn:qftLecture5:660}
\antisymmetric{\hat{\Pi}(\By, t)}{\tilde{\phi}(\Bx, t)} = -i \delta^3(\Bx – \By)

## Hamiltonian for a scalar field

In [1] it is left as an exersize to expand the scalar field Hamiltonian in terms of the raising and lowering operators. Let’s do that.

The field operator expanded in terms of the raising and lowering operators is

\label{eqn:scalarFieldHamiltonian:20}
\phi(x) =
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i k \cdot x}
+ a_k^\dagger e^{i k \cdot x}
}.

Note that $$x$$ and $$k$$ here are both four-vectors, so this field is dependent on a spacetime point, but the integration is over a spatial volume.

The Hamiltonian in terms of the fields was
\label{eqn:scalarFieldHamiltonian:40}
H = \inv{2} \int d^3 x \lr{ \Pi^2 + \lr{ \spacegrad \phi }^2 + \mu^2 \phi^2 }.

The field derivatives are

\label{eqn:scalarFieldHamiltonian:60}
\Pi
= \partial_0 \phi
= \partial_0
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}
=
i
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \frac{\omega_k}{ 2 \omega_k } } \lr{
-a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
},

and

\label{eqn:scalarFieldHamiltonian:80}
\partial_n \phi
= \partial_n
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}
=
i
\int \frac{ d^3 k k^n}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
-a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}.

Introducing a second set of momentum variables with $$j = \Abs{\Bj}$$, the momentum portion of the Hamiltonian is

\label{eqn:scalarFieldHamiltonian:100}
\begin{aligned}
\inv{2} \int d^3 x \Pi^2
&=
-\inv{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\omega_j
\omega_k
\lr{
-a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
+a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
-a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
-\inv{4}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\sqrt{
\omega_j
\omega_k}
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t – i (\Bk + \Bj) \cdot \Bx}
+ a_j a_k e^{-i (\omega_j + \omega_k) t + i (\Bj + \Bk) \cdot \Bx}
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t – i (\Bj – \Bk) \cdot \Bx}
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t – i (\Bk – \Bj) \cdot \Bx}
} \\
&=
-\inv{4}
\int
d^3 j
d^3 k
\sqrt{
\omega_j
\omega_k}
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t } \delta^3(\Bk + \Bj)
+ a_j a_k e^{-i (\omega_j + \omega_k) t } \delta^3(-\Bj – \Bk)
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t } \delta^3(\Bj – \Bk)
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t } \delta^3(\Bk – \Bj)
} \\
&=
-\inv{4}
\int
d^3 k
\omega_k
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}.
\end{aligned}

For the gradient portion of the Hamiltonian we have

\label{eqn:scalarFieldHamiltonian:120}
\begin{aligned}
\inv{2} \int d^3 x \lr{ \spacegrad \phi }^2
&=
-\inv{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{ \sum_{n=1}^3 j^n k^n }
\lr{
a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
-a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
-a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
-\inv{4}
\int
d^3 j
d^3 k
\frac{\Bj \cdot \Bk}{ \sqrt{ \omega_j \omega_k } }
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t } \delta^3(\Bk + \Bj)
+ a_j a_k e^{-i (\omega_j + \omega_k) t } \delta^3(-\Bj – \Bk)
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t } \delta^3(\Bj – \Bk)
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t } \delta^3(\Bk – \Bj)
} \\
&=
-\inv{4}
\int
d^3 k
\frac{\Bk^2}{ \omega_k }
\lr{
– a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
– a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}.
\end{aligned}

Finally, for the mass term, we have

\label{eqn:scalarFieldHamiltonian:140}
\begin{aligned}
\inv{2} \int d^3 x \mu^2 \phi^2
&=
\frac{\mu^2}{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
+a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
\frac{\mu^2}{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j a_k e^{-i (\omega_k + \omega_j) t + i (\Bk + \Bj) \cdot \Bx}
+a_j^\dagger a_k^\dagger e^{i (\omega_j + \omega_k) t – i (\Bk + \Bj) \cdot \Bx}
+a_j a_k^\dagger e^{i (\omega_k – \omega_j) t – i (\Bk – \Bj) \cdot \Bx}
+a_j^\dagger a_k e^{-i (\omega_k + \omega_j) t – i (\Bj – \Bk) \cdot \Bx}
} \\
&=
\frac{\mu^2}{2}
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j a_k e^{-i (\omega_k + \omega_j) t } \delta^3(- \Bk – \Bj)
+a_j^\dagger a_k^\dagger e^{i (\omega_j + \omega_k) t } \delta^3( \Bk + \Bj)
+a_j a_k^\dagger e^{i (\omega_k – \omega_j) t } \delta^3 (\Bk – \Bj)
+a_j^\dagger a_k e^{-i (\omega_k + \omega_j) t } \delta^3 (\Bj – \Bk)
} \\
&=
\frac{\mu^2}{4}
\int
d^3 k
\inv{ \omega_k }
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}.
\end{aligned}

Now all the pieces can be put back together again

\label{eqn:scalarFieldHamiltonian:160}
\begin{aligned}
H
&=
\inv{4}
\int d^3 k
\inv{\omega_k}
\lr{
-\omega_k^2
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}
+
\Bk^2
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
+ a_k^\dagger a_k
+ a_k a_k^\dagger
}
+
\mu^2
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}
} \\
&=
\inv{4}
\int d^3 k
\inv{\omega_k}
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
\lr{
-\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_{-k} a_k e^{- 2 i \omega_k t }
\lr{
-\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_k a_k^\dagger
\lr{
\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_k^\dagger a_k
\lr{
\omega_k^2
+ \Bk^2
+
\mu^2
}
}.
\end{aligned}

With $$\omega_k^2 = \Bk^2 + \mu^2$$, the time dependent terms are killed leaving
\label{eqn:scalarFieldHamiltonian:180}
H
=
\inv{2}
\int d^3 k
\omega_k
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.

# References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL lecturenotes.pdf. [Online; accessed 02-Jan-2016].

## Spin three halves spin interaction

### Q: [1] pr 3.33

A spin $$3/2$$ nucleus subjected to an external electric field has an interaction Hamiltonian of the form

\label{eqn:spinThreeHalvesNucleus:20}
H = \frac{e Q}{2 s(s-1) \Hbar^2} \lr{
\lr{\PDSq{x}{\phi}}_0 S_x^2
+\lr{\PDSq{y}{\phi}}_0 S_y^2
+\lr{\PDSq{z}{\phi}}_0 S_z^2
}.

Show that the interaction energy can be written as

\label{eqn:spinThreeHalvesNucleus:40}
A(3 S_z^2 – \BS^2) + B(S_{+}^2 + S_{-}^2).

Find the energy eigenvalues for such a Hamiltonian.

### A:

Reordering
\label{eqn:spinThreeHalvesNucleus:60}
\begin{aligned}
S_{+} &= S_x + i S_y \\
S_{-} &= S_x – i S_y,
\end{aligned}

gives
\label{eqn:spinThreeHalvesNucleus:80}
\begin{aligned}
S_x &= \inv{2} \lr{ S_{+} + S_{-} } \\
S_y &= \inv{2i} \lr{ S_{+} – S_{-} }.
\end{aligned}

The squared spin operators are
\label{eqn:spinThreeHalvesNucleus:100}
\begin{aligned}
S_x^2
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + S_{+} S_{-} + S_{-} S_{+} } \\
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + 2( S_x^2 + S_y^2 ) } \\
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + 2( \BS^2 – S_z^2 ) },
\end{aligned}

\label{eqn:spinThreeHalvesNucleus:120}
\begin{aligned}
S_y^2
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – S_{+} S_{-} – S_{-} S_{+} } \\
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – 2( S_x^2 + S_y^2 ) } \\
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – 2( \BS^2 – S_z^2 ) }.
\end{aligned}

This gives
\label{eqn:spinThreeHalvesNucleus:140}
\begin{aligned}
H &= \frac{e Q}{2 s(s-1) \Hbar^2} \biglr{ \inv{4} \lr{\PDSq{x}{\phi}}_0 \lr{ S_{+}^2 + S_{-}^2 + 2( \BS^2 – S_z^2 ) }
-\lr{\PDSq{y}{\phi}}_0 \lr{ S_{+}^2 + S_{-}^2 – 2( \BS^2 – S_z^2 ) }
+\lr{\PDSq{z}{\phi}}_0 S_z^2 } \\
&= \frac{e Q}{2 s(s-1) \Hbar^2} \biglr{ \inv{4} \lr{ \lr{\PDSq{x}{\phi}}_0 -\lr{\PDSq{y}{\phi}}_0 } \lr{ S_{+}^2 + S_{-}^2 }
+ \inv{2} \lr{ \lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0 } \BS^2
+ \lr{ \lr{\PDSq{z}{\phi}}_0 – \inv{2} \lr{\PDSq{x}{\phi}}_0 – \inv{2} \lr{\PDSq{y}{\phi}}_0 } S_z^2
}.
\end{aligned}

For a static electric field we have

\label{eqn:spinThreeHalvesNucleus:160}

but are evaluating it at a point away from the generating charge distribution, so $$\spacegrad^2 \phi = 0$$ at that point. This gives

\label{eqn:spinThreeHalvesNucleus:180}
H
=
\frac{e Q}{4 s(s-1) \Hbar^2}
\biglr{
\inv{2} \lr{ \lr{\PDSq{x}{\phi}}_0 -\lr{\PDSq{y}{\phi}}_0
} \lr{ S_{+}^2 + S_{-}^2 }
+
\lr{
\lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0
} (\BS^2 – 3 S_z^2)
},

so
\label{eqn:spinThreeHalvesNucleus:200}
A =
-\frac{e Q}{4 s(s-1) \Hbar^2} \lr{
\lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0
}

\label{eqn:spinThreeHalvesNucleus:220}
B =
\frac{e Q}{8 s(s-1) \Hbar^2}
\lr{ \lr{\PDSq{x}{\phi}}_0 – \lr{\PDSq{y}{\phi}}_0 }.

### A: energy eigenvalues

Using sakuraiProblem3.33.nb, matrix representations for the spin three halves operators and the Hamiltonian were constructed with respect to the basis $$\setlr{ \ket{3/2}, \ket{1/2}, \ket{-1/2}, \ket{-3/2} }$$

\label{eqn:spinThreeHalvesNucleus:240}
\begin{aligned}
S_{+} &=
\Hbar
\begin{bmatrix}
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & \sqrt{3} \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \\
S_{-} &=
\Hbar
\begin{bmatrix}
0 & 0 & 0 & 0 \\
\sqrt{3} & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & \sqrt{3} & 0 \\
\end{bmatrix} \\
S_x &=
\Hbar
\begin{bmatrix}
0 & \sqrt{3}/2 & 0 & 0 \\
\sqrt{3}/2 & 0 & 1 & 0 \\
0 & 1 & 0 & \sqrt{3}/2 \\
0 & 0 & \sqrt{3}/2 & 0 \\
\end{bmatrix} \\
S_y &=
i \Hbar
\begin{bmatrix}
0 & -\ifrac{\sqrt{3}}{2} & 0 & 0 \\
\ifrac{\sqrt{3}}{2} & 0 & -1 & 0 \\
0 & 1 & 0 & -\ifrac{\sqrt{3}}{2} \\
0 & 0 & \ifrac{\sqrt{3}}{2} & 0 \\
\end{bmatrix} \\
S_z &=
\frac{\Hbar}{2}
\begin{bmatrix}
3 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -3 \\
\end{bmatrix} \\
H &=
\begin{bmatrix}
3 A & 0 & 2 \sqrt{3} B & 0 \\
0 & -3 A & 0 & 2 \sqrt{3} B \\
2 \sqrt{3} B & 0 & -3 A & 0 \\
0 & 2 \sqrt{3} B & 0 & 3 A \\
\end{bmatrix}.
\end{aligned}

The energy eigenvalues are found to be

\label{eqn:spinThreeHalvesNucleus:260}
E = \pm \Hbar^2 \sqrt{9 A^2 + 12 B^2 },

with two fold degeneracies for each eigenvalue.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## PHY1520H Graduate Quantum Mechanics. Lecture 14: Angular momentum (cont.). Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 3 content.

### Review: Angular momentum

Given eigenket $$\ket{a, b}$$, where

\label{eqn:qmLecture14:20}
\begin{aligned}
\hat{\BL}^2 \ket{a, b} &= \Hbar^2 a \ket{a,b} \\
\hat{L}_z \ket{a, b} &= \Hbar b \ket{a,b}
\end{aligned}

We were looking for

\label{eqn:qmLecture14:40}
\hat{L}_{x,y} \ket{a,b} = \sum_{b’} \mathcal{A}^{x,y}_{a; b, b’} \ket{a,b’},

by applying

\label{eqn:qmLecture14:60}
\hat{L}_{\pm} = \hat{L}_x \pm i \hat{L}_y.

We found

\label{eqn:qmLecture14:80}
\hat{L}_{\pm} \propto \ket{a, b \pm 1}.

Let

\label{eqn:qmLecture14:100}
\ket{\phi_\pm} = \hat{L}_{\pm} \ket{a, b}.

We want

\label{eqn:qmLecture14:120}
\braket{\phi_\pm}{\phi_\pm} \ge 0,

or
\label{eqn:qmLecture14:140}
\begin{aligned}
\bra{a,b} \hat{L}_{+} \hat{L}_{-} \ket{a, b} &\ge 0 \\
\bra{a,b} \hat{L}_{-} \hat{L}_{+} \ket{a, b} &\ge 0
\end{aligned}

We found

\label{eqn:qmLecture14:160}
\begin{aligned}
\hat{L}_{+} \hat{L}_{-} =
\lr{ \hat{L}_x + i \hat{L}_y } \lr{ \hat{L}_x – i \hat{L}_y }
&= \lr{ \hat{\BL}^2 – \hat{L}_z^2 } -i \antisymmetric{\hat{L}_x}{\hat{L}_y} \\
&= \lr{ \hat{\BL}^2 – \hat{L}_z^2 } -i \lr{ i \Hbar \hat{L}_z } \\
&= \lr{ \hat{\BL}^2 – \hat{L}_z^2 } + \Hbar \hat{L}_z,
\end{aligned}

so

\label{eqn:qmLecture14:180}
\bra{a,b} \hat{L}_{+} \hat{L}_{-} \ket{a, b}
=
\expectation{ \hat{\BL}^2 – \hat{L}_z^2 + \Hbar \hat{L}_z }.

Similarly
\label{eqn:qmLecture14:200}
\bra{a,b} \hat{L}_{-} \hat{L}_{+} \ket{a, b}
=
\expectation{ \hat{\BL}^2 – \hat{L}_z^2 – \Hbar \hat{L}_z }.

### Constraints

\label{eqn:qmLecture14:220}
\begin{aligned}
a – b^2 + b &\ge 0 \\
a – b^2 – b &\ge 0
\end{aligned}

If these are satisfied at the equality extreme we have

\label{eqn:qmLecture14:240}
\begin{aligned}
b_{\textrm{max}} \lr{ b_{\textrm{max}} + 1 } &= a \\
b_{\textrm{min}} \lr{ b_{\textrm{min}} – 1 } &= a.
\end{aligned}

Rearranging this to solve, we can rewrite the equality as

\label{eqn:qmLecture14:680}
\lr{ b_{\textrm{max}} + \inv{2} }^2 – \inv{4} = \lr{ b_{\textrm{min}} – \inv{2} }^2 – \inv{4},

which has solutions at

\label{eqn:qmLecture14:700}
b_{\textrm{max}} + \inv{2} = \pm \lr{ b_{\textrm{min}} – \inv{2} }.

One of the solutions is

\label{eqn:qmLecture14:260}
-b_{\textrm{min}} = b_{\textrm{max}}.

The other solution is $$b_{\textrm{max}} = b_{\textrm{min}} – 1$$, which we discard.

The final constraint is therefore

\label{eqn:qmLecture14:280}
\boxed{
– b_{\textrm{max}} \le b \le b_{\textrm{max}},
}

and

\label{eqn:qmLecture14:320}
\begin{aligned}
\hat{L}_{+} \ket{a, b_{\textrm{max}}} &= 0 \\
\hat{L}_{-} \ket{a, b_{\textrm{min}}} &= 0
\end{aligned}

If we had the sequence, which must terminate at $$b_{\textrm{min}}$$ or else it will go on forever

\label{eqn:qmLecture14:340}
\ket{a, b_{\textrm{max}}}
\overset{\hat{L}_{-}}{\rightarrow}
\ket{a, b_{\textrm{max}} – 1}
\overset{\hat{L}_{-}}{\rightarrow}
\ket{a, b_{\textrm{max}} – 2}
\cdots
\overset{\hat{L}_{-}}{\rightarrow}
\ket{a, b_{\textrm{min}}},

then we know that $$b_{\textrm{max}} – b_{\textrm{min}} \in \mathbb{Z}$$, or

\label{eqn:qmLecture14:360}
b_{\textrm{max}} – n = b_{\textrm{min}} = -b_{\textrm{max}}

or

\label{eqn:qmLecture14:380}
b_{\textrm{max}} = \frac{n}{2},

this is either an integer or a $$1/2$$ odd integer, depending on whether $$n$$ is even or odd. These are called “orbital” or “spin” respectively.

The convention is to write

\label{eqn:qmLecture14:400}
\begin{aligned}
b_{\textrm{max}} &= j \\
a &= j(j + 1).
\end{aligned}

so for $$m \in -j, -j + 1, \cdots, +j$$

\label{eqn:qmLecture14:420}
\boxed{
\begin{aligned}
\hat{\BL}^2 \ket{j, m} &= \Hbar^2 j (j + 1) \ket{j, m} \\
L_z \ket{j, m} &= \Hbar m \ket{j, m}.
\end{aligned}
}

## Schwinger’s Harmonic oscillator representation of angular momentum operators.

In [2] a powerful method for describing angular momentum with harmonic oscillators was introduced, which will be outlined here. The question is whether we can construct a set of harmonic oscillators that allows a mapping from

\label{eqn:qmLecture14:460}
\hat{L}_{+} \leftrightarrow a^{+}?

Picture two harmonic oscillators, one with states counted from one zero towards $$\infty$$ and another with states counted from a different zero towards $$-\infty$$, as pictured in fig. 1.

fig. 1. Overlapping SHO domains

Is it possible that such an overlapping set of harmonic oscillators can provide the properties of the angular momentum operators? Let’s relabel the counting so that we have two sets of positive counted SHO systems, each counted in a positive direction as sketched in fig. 2.

fig. 2. Relabeling the counting for overlapping SHO systems

It turns out that given a constraint there the number of ways to distribute particles between a pair of SHO systems, the process that can be viewed as reproducing the angular momentum action is a transfer of particles from one harmonic oscillator to the other. For $$\hat{L}_z = +j$$

\label{eqn:qmLecture14:480}
\begin{aligned}
n_1 &= n_{\textrm{max}} \\
n_2 &= 0,
\end{aligned}

and for $$\hat{L}_z = -j$$

\label{eqn:qmLecture14:500}
\begin{aligned}
n_1 &= 0 \\
n_2 &= n_{\textrm{max}}.
\end{aligned}

We can make the identifications

\label{eqn:qmLecture14:520}
\hat{L}_z = \lr{ n_1 – n_2 } \frac{\Hbar}{2},

and
\label{eqn:qmLecture14:540}
j = \inv{2} n_{\textrm{max}},

or

\label{eqn:qmLecture14:560}
n_1 + n_2 = \text{fixed} = n_{\textrm{max}}

Changes that keep $$n_1 + n_2$$ fixed are those that change $$n_1$$, $$n_2$$ by $$+1$$ or $$-1$$ respectively, as sketched in fig. 3.

fig. 3. Number conservation constraint.

Can we make an identification that takes

\label{eqn:qmLecture14:580}
\ket{n_1, n_2} \overset{\hat{L}_{-}}{\rightarrow} \ket{n_1 – 1, n_2 + 1}?

What operator in the SHO problem has this effect? Let’s try

\boxedEquation{eqn:qmLecture14:620}{
\begin{aligned}
\hat{L}_{-} &= \Hbar a_2^\dagger a_1 \\
\hat{L}_{+} &= \Hbar a_1^\dagger a_2 \\
\hat{L}_z &= \frac{\Hbar}{2} \lr{ n_1 – n_2 }
\end{aligned}
}

Is this correct? Do we need to make any scalar adjustments? We want

\label{eqn:qmLecture14:640}
\antisymmetric{\hat{L}_z}{\hat{L}_{\pm}} = \pm \Hbar \hat{L}_{\pm}.

First check this with the $$\hat{L}_{+}$$ commutator

\label{eqn:qmLecture14:660}
\begin{aligned}
\antisymmetric{\hat{L}_z}{\hat{L}_{+}}
&=
\inv{2} \Hbar^2 \antisymmetric{ n_1 – n_2}{a_1^\dagger a_2 } \\
&=
\inv{2} \Hbar^2 \antisymmetric{ a_1^\dagger a_1 – a_2^\dagger a_2 }{a_1^\dagger a_2 } \\
&=
\inv{2} \Hbar^2
\lr{
\antisymmetric{ a_1^\dagger a_1 }{a_1^\dagger a_2 }
-\antisymmetric{ a_2^\dagger a_2 }{a_1^\dagger a_2 }
} \\
&=
\inv{2} \Hbar^2
\lr{
a_2 \antisymmetric{ a_1^\dagger a_1 }{a_1^\dagger }
-a_1^\dagger \antisymmetric{ a_2^\dagger a_2 }{a_2 }
}.
\end{aligned}

But

\label{eqn:qmLecture14:720}
\begin{aligned}
\antisymmetric{ a^\dagger a }{a^\dagger }
&=
a^\dagger a
a^\dagger

a^\dagger
a^\dagger a \\
&=
a^\dagger \lr{ 1 +
a^\dagger a}

a^\dagger
a^\dagger a \\
&=
a^\dagger,
\end{aligned}

and
\label{eqn:qmLecture14:740}
\begin{aligned}
\antisymmetric{ a^\dagger a }{a}
&=
a^\dagger a a
-a a^\dagger a \\
&=
a^\dagger a a
-\lr{ 1 + a^\dagger a } a \\
&=
-a,
\end{aligned}

so
\label{eqn:qmLecture14:760}
\antisymmetric{\hat{L}_z}{\hat{L}_{+}} = \Hbar^2 a_2 a_1^\dagger = \Hbar \hat{L}_{+},

as desired. Similarly

\label{eqn:qmLecture14:780}
\begin{aligned}
\antisymmetric{\hat{L}_z}{\hat{L}_{-}}
&=
\inv{2} \Hbar^2 \antisymmetric{ n_1 – n_2}{a_2^\dagger a_1 } \\
&=
\inv{2} \Hbar^2 \antisymmetric{ a_1^\dagger a_1 – a_2^\dagger a_2 }{a_2^\dagger a_1 } \\
&=
\inv{2} \Hbar^2 \lr{
a_2^\dagger \antisymmetric{ a_1^\dagger a_1 }{a_1 }
– a_1 \antisymmetric{ a_2^\dagger a_2 }{a_2^\dagger }
} \\
&=
\inv{2} \Hbar^2 \lr{
a_2^\dagger (-a_1)
– a_1 a_2^\dagger
} \\
&=
– \Hbar^2 a_2^\dagger a_1 \\
&=
– \Hbar \hat{L}_{-}.
\end{aligned}

With

\label{eqn:qmLecture14:800}
\begin{aligned}
j &= \frac{n_1 + n_2}{2} \\
m &= \frac{n_1 – n_2}{2} \\
\end{aligned}

We can make the identification

\label{eqn:qmLecture14:820}
\ket{n_1, n_2} = \ket{ j+ m , j – m}.

### Another way

With

\label{eqn:qmLecture14:840}
\hat{L}_{+} \ket{j, m} = d_{j,m}^{+} \ket{j, m+1}

or

\label{eqn:qmLecture14:860}
\Hbar a_1^\dagger a_2 \ket{j + m, j-m} = d_{j,m}^{+} \ket{ j + m + 1, j- m-1},

we can seek this factor $$d_{j,m}^{+}$$ by operating with $$\hat{L}_{+}$$

\label{eqn:qmLecture14:880}
\begin{aligned}
\hat{L}_{+} \ket{j, m}
&=
\Hbar a_1^\dagger a_2 \ket{n_1, n_2} \\
&=
\Hbar a_1^\dagger a_2 \ket{j+m,j-m} \\
&=
\Hbar \sqrt{ n + 1 } \sqrt{n_2} \ket{j+m +1,j-m-1} \\
&=
\Hbar \sqrt{ \lr{ j+ m + 1}\lr{ j – m } } \ket{j+m +1,j-m-1}
\end{aligned}

That gives
\label{eqn:qmLecture14:900}
\begin{aligned}
d_{j,m}^{+} &= \Hbar \sqrt{\lr{ j – m } \lr{ j+ m + 1} } \\
d_{j,m}^{-} &= \Hbar \sqrt{\lr{ j + m } \lr{ j- m + 1} }.
\end{aligned}

This equivalence can be used to model spin interaction in crystals as harmonic oscillators. This equivalence of lattice vibrations and spin oscillations is called “spin waves”.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

[2] J Schwinger. Quantum theory of angular momentum. biedenharn l., van dam h., editors, 1955. URL http://www.ifi.unicamp.br/ cabrera/teaching/paper_schwinger.pdf.

## PHY1520H Graduate Quantum Mechanics. Lecture 13: Time reversal (cont.), and angular momentum. Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering \textchapref{{4}}, \textchapref{{3}} [1] content.

## Time reversal (cont.)

Given a time reversed state

\label{eqn:qmLecture13:20}
\ket{\tilde{\Psi}(t)} = \Theta \ket{\Psi(0)}

which can alternately be written

\label{eqn:qmLecture13:40}
\Theta^{-1} \ket{\tilde{\Psi}(t)} = \ket{\Psi(-t)} = e^{i \hat{H} t/\Hbar} \ket{\Psi(0)}

The left hand side can be expanded as the evolution of the state as found at time $$-t$$

\label{eqn:qmLecture13:60}
\begin{aligned}
\Theta^{-1} \ket{\tilde{\Psi}(t)}
&=
\Theta^{-1} e^{-i \hat{H} t/\Hbar} \ket{\tilde{\Psi}(-t)} \\
&=
\Theta^{-1} e^{-i \hat{H} t/\Hbar} \Theta \ket{\Psi(0)}.
\end{aligned}

To first order for a small time increment $$\delta t$$, we have

\label{eqn:qmLecture13:80}
\lr{ 1 + i \frac{\hat{H}}{\Hbar} \delta t } \ket{\Psi(0)} =
\Theta^{-1} \lr{ 1 – i \frac{\hat{H}}{\Hbar} \delta t } \Theta \ket{\Psi(0)},

or

\label{eqn:qmLecture13:120}
i \frac{\hat{H}}{\Hbar} \delta t \ket{\Psi(0)}
=
\Theta^{-1} (- i) \frac{\hat{H}}{\Hbar} \delta t \Theta \ket{\Psi(0)}.

Since this holds for any state $$\ket{\Psi(0)}$$, the time reversal operator satisfies

\label{eqn:qmLecture13:140}
i \hat{H}
=
\Theta^{-1} (- i) \hat{H} \Theta.

Note that the factors of $$i$$ have not been canceled on purpose, since we are allowing for the time reversal operator to not necessarily commute with imaginary numbers.

There are two possible solutions

• If $$\Theta$$ is unitary where $$\Theta i = i \Theta$$, then

\label{eqn:qmLecture13:160}
\hat{H}
=
-\Theta^{-1} \hat{H} \Theta,

or
\label{eqn:qmLecture13:180}
\Theta \hat{H}
=
– \hat{H} \Theta.

Consider the implications of this on energy eigenstates
\label{eqn:qmLecture13:200}
\hat{H} \ket{\Psi_n} = E_n \ket{\Psi_n},

\label{eqn:qmLecture13:220}
\Theta \hat{H} \ket{\Psi_n} = E_n \Theta \ket{\Psi_n},

but

\label{eqn:qmLecture13:240}
-\hat{H} \Theta \ket{\Psi_n} = E_n \Theta \ket{\Psi_n},

or

\label{eqn:qmLecture13:260}
\hat{H} \lr{ \Theta \ket{\Psi_n}} = -E_n \lr{ \Theta \ket{\Psi_n} }.

This would mean that $$\lr{ \Theta \ket{\Psi_n}}$$ is an eigenket of $$\hat{H}$$, but with a negative energy eigenvalue.

• $$\Theta$$ is antiunitary, where $$\Theta i = -i \Theta$$.

This time
\label{eqn:qmLecture13:280}
i \hat{H} = i \Theta^{-1} \hat{H} \Theta,

so

\label{eqn:qmLecture13:300}
\Theta \hat{H} = \hat{H} \Theta.

Acting on an energy eigenket, we’ve got

\label{eqn:qmLecture13:1400}
\Theta \hat{H} \ket{\Psi_n}
=
E_n \lr{ \Theta \ket{\Psi_n} },

and
\label{eqn:qmLecture13:1420}
\lr{ \hat{H} \Theta } \ket{\Psi_n}
=
\hat{H} \lr{ \Theta \ket{\Psi_n} },

so $$\Theta \ket{\Psi_n}$$ is an eigenstate with energy $$E_n$$.

### What properties do we expect from $$\Theta$$?

We expect
\label{eqn:qmLecture13:320}
\begin{aligned}
\hat{x} &\rightarrow \hat{x} \\
\hat{p} &\rightarrow -\hat{p} \\
\hat{\BL} &\rightarrow -\hat{\BL}
\end{aligned}

where we have a sign flip in the time dependent momentum operator (and therefore angular momentum), but not for position. If we have

\label{eqn:qmLecture13:340}
\Theta^{-1} \hat{x} \Theta = \hat{x},

if that’s true, then how about the momentum operator in the position basis
\label{eqn:qmLecture13:360}
\begin{aligned}
\Theta^{-1} \hat{p} \Theta
&=
\Theta^{-1} \lr{ -i \Hbar \PD{x}{} } \Theta \\
&=
\Theta^{-1} \lr{ -i \Hbar } \Theta \PD{x}{} \\
&=
i \Hbar \Theta^{-1} \Theta \PD{x}{} \\
&=
-\hat{p}.
\end{aligned}

How about the $$x,p$$ commutator? For that we have

\label{eqn:qmLecture13:380}
\begin{aligned}
\Theta^{-1} \antisymmetric{\hat{x}}{\hat{p}} \Theta
&=
\Theta^{-1} \lr{ i \Hbar } \Theta \\
&=
-i \Hbar \Theta^{-1} \Theta \\
&=
– \antisymmetric{\hat{x}}{\hat{p}}.
\end{aligned}

For the the angular momentum operators

\label{eqn:qmLecture13:420}
\hat{L}_i = \epsilon_{ijk} \hat{r}_j \hat{p}_k,

the time reversal operator should flip the sign due to its action on $$\hat{p}_k$$.

### Time reversal acting on spin 1/2 (Fermions). Attempt I.

Consider two spin states $$\ket{\uparrow}, \ket{\downarrow}$$. What should the action of the time reversal operator on such a state be? Let’s (incorrectly) start by supposing that the time reversal operator effects are

\label{eqn:qmLecture13:440}
\begin{aligned}
\Theta \ket{\uparrow} &= \ket{\downarrow} \\
\Theta \ket{\downarrow} &= \ket{\uparrow}.
\end{aligned}

Given a general state
so that if

\label{eqn:qmLecture13:740}
\ket{\Psi} = a \ket{\uparrow} + b \ket{\downarrow},

the action of the time reversal operator would be

\label{eqn:qmLecture13:760}
\Theta \ket{\Psi} = a^\conj \ket{\downarrow} + b^\conj \ket{\uparrow}.

That action is:

\label{eqn:qmLecture13:460}
\begin{aligned}
a \rightarrow b^\conj \\
b \rightarrow a^\conj
\end{aligned}

Let’s consider whether or not such an action a spin operator with properties

\label{eqn:qmLecture13:480}
\antisymmetric{\hat{S}_i}{\hat{S}_j} = i \epsilon_{ijk} \hat{S}_k.

produce the desired inversion of sign

\label{eqn:qmLecture13:500}
\Theta^{-1} \hat{S}_i \Theta = – \hat{S}_i.

The expectations of the spin operators (without any application of time reversal) are

\label{eqn:qmLecture13:1440}
\begin{aligned}
\bra{\Psi} \hat{S}_x \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\sigma_x
\lr{ a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj b + b^\conj a },
\end{aligned}

\label{eqn:qmLecture13:1460}
\begin{aligned}
\bra{\Psi} \hat{S}_y \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\sigma_y
\lr{ a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{i\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\lr{ a \ket{\downarrow} – b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2 i} \lr{ a^\conj b – b^\conj a },
\end{aligned}

\label{eqn:qmLecture13:1480}
\begin{aligned}
\bra{\Psi} \hat{S}_z \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\sigma_z
\lr{ a \ket{\uparrow} – b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2} \lr{ \Abs{a}^2 – \Abs{b}^2 }
\end{aligned}

The time reversed actions are

\label{eqn:qmLecture13:1560}
\begin{aligned}
\bra{\Psi} \Theta^{-1} \hat{S}_x \Theta \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\sigma_x
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\lr{ a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj b + b^\conj a },
\end{aligned}

\label{eqn:qmLecture13:1580}
\begin{aligned}
\bra{\Psi} \Theta^{-1} \hat{S}_y \Theta \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\sigma_y
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{i\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\lr{ -a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2 i} \lr{ -a^\conj b + b^\conj a },
\end{aligned}

\label{eqn:qmLecture13:1600}
\begin{aligned}
\bra{\Psi} \Theta^{-1} \hat{S}_z \Theta \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\sigma_z
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\lr{ -a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2} \lr{ -\Abs{a}^2 + \Abs{b}^2 }
\end{aligned}

We see that this is not right, because the sign for the x component has not been flipped.

### Spin 1/2 (Fermions). Attempt II.

Again assuming

\label{eqn:qmLecture13:580}
\ket{\Psi} = a \ket{\uparrow} + b \ket{\downarrow},

now try the action

\label{eqn:qmLecture13:780}
\Theta \ket{\Psi} = a^\conj \ket{\downarrow} – b^\conj \ket{\uparrow}.

This is the action:

\label{eqn:qmLecture13:600}
\begin{aligned}
a \rightarrow -b^\conj \\
b \rightarrow a^\conj
\end{aligned}

The correct action of time reversal on the basis states (up to a phase choice) is

\label{eqn:qmLecture13:630}
\boxed{
\begin{aligned}
\Theta \ket{\uparrow} &= \ket{\downarrow} \\
\Theta \ket{\downarrow} &= -\ket{\uparrow} \\
\end{aligned}
}

Note that acting the time reversal operator twice has the effects

\label{eqn:qmLecture13:660}
\Theta^2 \ket{\uparrow} = \Theta \ket{\downarrow} = – \ket{\uparrow}

\label{eqn:qmLecture13:680}
\Theta^2 \ket{\downarrow} = \Theta (-\ket{\uparrow}) = – \ket{\uparrow}.

We end up with the same state we started with, but with the opposite sign. This means that as an operator

\label{eqn:qmLecture13:700}
\boxed{
\Theta^2 = -1.
}

This is try for half integer particles (Fermions) $$S = 1/2, 3/2, 5/2, \cdots$$, but for Bosons with integer spin $$S$$.

\label{eqn:qmLecture13:720}
\boxed{
\Theta^2 = 1.
}

### Kramer’s degeneracy for Spin 1/2 (Fermions)

Suppose we imagine there is state for which the action of the time reversal operator products the same state, just different in phase

\label{eqn:qmLecture13:800}
\begin{aligned}
\Theta \ket{\Psi_n}
&= \ket{\tilde{\Psi}_n} \\
&= e^{i \delta} \ket{\tilde{\Psi}_n},
\end{aligned}

then
\label{eqn:qmLecture13:840}
\begin{aligned}
\Theta^2 \ket{\Psi_n}
&= \Theta e^{i \delta} \ket{\tilde{\Psi}_n} \\
&= e^{i \delta} e^{i \delta} \ket{\tilde{\Psi}_n},
\end{aligned}

but

\label{eqn:qmLecture13:860}
\begin{aligned}
\Theta e^{i \delta} \ket{\tilde{\Psi}_n}
&=
e^{-i \delta} \Theta \ket{\tilde{\Psi}_n} \\
&=
e^{-i \delta} e^{i \delta} \ket{\tilde{\Psi}_n} \\
&=
\ket{\tilde{\Psi}_n}
\ne
– \ket{\tilde{\Psi}_n}.
\end{aligned}

This is a contradiction, so we must have at least a two-fold degeneracy. This is called Kramer’s degeneracy. In the homework we will show that this is not the case for integer spin particles.

## Angular momentum

In classical mechanics the (orbital) angular momentum is

\label{eqn:qmLecture13:880}
\BL = \Br \cross \Bp.

Here “orbital” is to distinguish from spin angular momentum.

In quantum mechanics, the mapping to operators, in component form, is

\label{eqn:qmLecture13:900}
\hat{L}_i = \epsilon_{ijk} \hat{r}_j \hat{p}_k.

These operators do not commute
\label{eqn:qmLecture13:920}
\antisymmetric{\hat{L}_i}{\hat{L}_j}
=
i \Hbar \epsilon_{ijk} \hat{L}_k.

which means that we can’t simultaneously determine $$\hat{L}_i$$ for all $$i$$.

Aside: In quantum mechanics, we define an operator $$\hat{\BV}$$ to be a vector operator if

\label{eqn:qmLecture13:940}
\antisymmetric{\hat{L}_i}{\hatV_j}
=
i \Hbar \epsilon_{ijk} \hatV_k.

The commutator of the squared angular momentum operator with any $$\hat{L}_i$$, say $$\hat{L}_x$$ is zero

\label{eqn:qmLecture13:960}
\begin{aligned}
\antisymmetric{
\hat{L}_x^2 +
\hat{L}_y^2 +
\hat{L}_z^2
}
{\hat{L}_x}
&=
\hat{L}_y \hat{L}_y \hat{L}_x
– \hat{L}_x \hat{L}_y \hat{L}_y
+
\hat{L}_z \hat{L}_z \hat{L}_x
– \hat{L}_x \hat{L}_z \hat{L}_z \\
&=
\hat{L}_y \lr{ \antisymmetric{\hat{L}_y}{\hat{L}_x} + {\hat{L}_x \hat{L}_y} }
-\lr{ \antisymmetric{\hat{L}_x}{\hat{L}_y} + {\hat{L}_y \hat{L}_x} } \hat{L}_y \\
&\quad +\hat{L}_z \lr{ \antisymmetric{\hat{L}_z}{\hat{L}_x} + {\hat{L}_x \hat{L}_z} }
-\lr{ \antisymmetric{\hat{L}_x}{\hat{L}_z} + {\hat{L}_z \hat{L}_x} } \hat{L}_z \\
&=
\hat{L}_y \antisymmetric{\hat{L}_y}{\hat{L}_x}
-\antisymmetric{\hat{L}_x}{\hat{L}_y} \hat{L}_y
+\hat{L}_z \antisymmetric{\hat{L}_z}{\hat{L}_x}
-\antisymmetric{\hat{L}_x}{\hat{L}_z} \hat{L}_z \\
&=
i \Hbar \lr{
-\hat{L}_y \hat{L}_z
– \hat{L}_z \hat{L}_y
+\hat{L}_z \hat{L}_y
+ \hat{L}_y \hat{L}_z
} \\
&=
0.
\end{aligned}

Suppose we have a state $$\ket{\Psi}$$ with a well defined $$\hat{L}_z$$ eigenvalue and well defined $$\hat{\BL^2}$$ eigenvalue, written as

\label{eqn:qmLecture13:1000}
\ket{\Psi} = \ket{a, b},

where the label $$a$$ is used for the eigenvalue of $$\hat{\BL}^2$$ and $$b$$ labels the eigenvalue of $$\hat{L}_z$$. Then

\label{eqn:qmLecture13:1020}
\begin{aligned}
\hat{\BL}^2 \ket{a , b} &= \Hbar^2 a \ket{a ,b} \\
\hat{L}_z \ket{a , b} &= \Hbar b \ket{a ,b}.
\end{aligned}

Things aren’t so nice when we act with other angular momentum operators, producing a scrambled mess

\label{eqn:qmLecture13:1040}
\begin{aligned}
\hat{L}_x \ket{a , b} &= \sum_{a’, b’} \mathcal{A}^x_{a, b, a’, b’} \ket{a’, b’} \\
\hat{L}_y \ket{a , b} &= \sum_{a’, b’} \mathcal{A}^y_{a, b, a’, b’} \ket{a’, b’} \\
\end{aligned}

With this representation, we have

\label{eqn:qmLecture13:1060}
\hat{L}_x \hat{\BL}^2 \ket{a, b}
=
\hat{L}_x \Hbar^2 a
\sum_{a’, b’} \mathcal{A}^x_{a, b, a’, b’} \ket{a’, b’}.

\label{eqn:qmLecture13:1080}
\hat{\BL}^2 \hat{L}_x \ket{a, b}
=
\Hbar^2
\sum_{a’, b’} a’ \mathcal{A}^x_{a, b, a’, b’} \ket{a’, b’}.

Since $$\hat{\BL}^2, \hat{L}_x$$ commute, we must have

\label{eqn:qmLecture13:1100}
\mathcal{A}^x_{a, b, a’, b’} = \delta_{a, a’} \mathcal{A}^x_{a’; b, b’},

and similarly
\label{eqn:qmLecture13:1120}
\mathcal{A}^y_{a, b, a’, b’} = \delta_{a, a’} \mathcal{A}^y_{a’; b, b’}.

Simplifying things we can write the action of $$\hat{L}_x, \hat{L}_y$$ on the state as

\label{eqn:qmLecture13:1140}
\begin{aligned}
\hat{L}_x \ket{a , b} &= \sum_{ b’} \mathcal{A}^x_{a; b, b’} \ket{a, b’} \\
\hat{L}_y \ket{a , b} &= \sum_{ b’} \mathcal{A}^y_{a; b, b’} \ket{a, b’} \\
\end{aligned}

Let’s define
\label{eqn:qmLecture13:1160}
\begin{aligned}
\hat{L}_{+} &\equiv \hat{L}_x + i \hat{L}_y \\
\hat{L}_{-} &\equiv \hat{L}_x – i \hat{L}_y \\
\end{aligned}

Because these are sums of $$\hat{L}_x, \hat{L}_y$$ they must also commute with $$\hat{\BL}^2$$

\label{eqn:qmLecture13:1180}
\antisymmetric{\hat{\BL}^2}{\hat{L}_{\pm}} = 0.

The commutators with $$\hat{L}_z$$ are non-zero

\label{eqn:qmLecture13:1740}
\begin{aligned}
\antisymmetric{\hat{L}_z}{\hat{L}_{\pm}}
&=
\hat{L}_z \lr{ \hat{L}_x \pm i \hat{L}_y }
– \lr{ \hat{L}_x \pm i \hat{L}_y } \hat{L}_z \\
&=
\antisymmetric{\hat{L}_z}{\hat{L}_x}
\pm i
\antisymmetric{\hat{L}_z}{\hat{L}_y} \\
&=
i \Hbar \lr{
\hat{L}_y \mp i \hat{L}_x
} \\
&=
\Hbar \lr{ i \hat{L}_y \pm \hat{L}_x } \\
&=
\pm \Hbar \lr{ \hat{L}_x \pm i \hat{L}_y } \\
&=
\pm \Hbar \hat{L}_{\pm}.
\end{aligned}

Explicitly, that is

\label{eqn:qmLecture13:1220}
\begin{aligned}
\hat{L}_z \hat{L}_{+} – \hat{L}_{+} \hat{L}_z &= \Hbar \hat{L}_{+} \\
\hat{L}_z \hat{L}_{-} – \hat{L}_{-} \hat{L}_z &= -\Hbar \hat{L}_{-}
\end{aligned}

Now we are set to compute actions of these (assumed) raising and lowering operators on the eigenstate of $$\hat{L}_z, \hat{\BL}^2$$

\label{eqn:qmLecture13:1240}
\begin{aligned}
\hat{L}_z \hat{L}_{\pm} \ket{a, b}
&=
\Hbar \hat{L}_{\pm} \ket{a,b} \pm \hat{L}_{\pm} \hat{L}_z \ket{a,b} \\
&=
\Hbar \hat{L}_{\pm} \ket{a,b} \pm \Hbar b \hat{L}_{\pm} \ket{a,b} \\
&=
\Hbar \lr{ b \pm 1 } \hat{L}_{\pm} \ket{a, b} .
\end{aligned}

There must be a proportionality of the form

\label{eqn:qmLecture13:1260}
\ket{\hat{L}_{\pm}} \propto \ket{a, b \pm 1},

The products of the raising and lowering operators are

\label{eqn:qmLecture13:1280}
\begin{aligned}
\hat{L}_{-} \hat{L}_{+}
&=
\lr{ \hat{L}_x – i \hat{L}_y }
\lr{ \hat{L}_x + i \hat{L}_y } \\
&=
\hat{L}_x^2 + \hat{L}_y^2 + i \hat{L}_x \hat{L}_y – i \hat{L}_y \hat{L}_x \\
&=
\lr{ \hat{\BL}^2 – \hat{L}_z^2 } + i \antisymmetric{\hat{L}_x}{\hat{L}_y} \\
&=
\hat{\BL}^2 – \hat{L}_z^2 – \Hbar \hat{L}_z,
\end{aligned}

and
\label{eqn:qmLecture13:1300}
\begin{aligned}
\hat{L}_{+} \hat{L}_{-}
&=
\lr{ \hat{L}_x + i \hat{L}_y }
\lr{ \hat{L}_x – i \hat{L}_y } \\
&=
\hat{L}_x^2 + \hat{L}_y^2 – i \hat{L}_x \hat{L}_y + i \hat{L}_y \hat{L}_x \\
&=
\lr{ \hat{\BL}^2 – \hat{L}_z^2 } – i \antisymmetric{\hat{L}_x}{\hat{L}_y} \\
&=
\hat{\BL}^2 – \hat{L}_z^2 + \Hbar \hat{L}_z,
\end{aligned}

So we must have

\label{eqn:qmLecture13:1320}
\begin{aligned}
0
&\le \bra{a, b} \hat{L}_{-} \hat{L}_{+} \ket{a, b} \\
&=
\bra{a, b}
\lr{ \hat{\BL}^2 – \hat{L}_z^2 – \Hbar \hat{L}_z }
\ket{a, b} \\
&=
\Hbar^2 a – \Hbar^2 b^2 – \Hbar^2 b,
\end{aligned}

and

\label{eqn:qmLecture13:1340}
\begin{aligned}
0
&\le \bra{a, b} \hat{L}_{+} \hat{L}_{-} \ket{a, b} \\
&=
\bra{a, b}
\lr{ \hat{\BL}^2 – \hat{L}_z^2 + \Hbar \hat{L}_z }
\ket{a, b} \\
&=
\Hbar^2 a – \Hbar^2 b^2 + \Hbar^2 b.
\end{aligned}

This puts constraints on $$a, b$$, roughly of the form

1. \label{eqn:qmLecture13:1360}
a – b( b + 1) \ge 0

With $$b_{\textrm{max}} > 0$$, $$b_{\textrm{max}} \approx \sqrt{a}$$.

2. \label{eqn:qmLecture13:1380}
a – b( b – 1) \ge 0

With $$b_{\textrm{min}} < 0$$, $$b_{\textrm{max}} \approx -\sqrt{a}$$.

## Question: Angular momentum commutators

Using $$\hat{L}_i = \epsilon_{ijk} \hat{r}_j \hat{p}_k$$, show that

\label{eqn:qmLecture13:1620}
\antisymmetric{\hat{L}_i}{\hat{L}_j} = i \Hbar \epsilon_{ijk} \hat{L}_k

Let’s start without using abstract index expressions, computing the commutator for $$\hat{L}_1, \hat{L}_2$$, which should show the basic steps required

\label{eqn:qmLecture13:1640}
\begin{aligned}
\antisymmetric{\hat{L}_1}{\hat{L}_2}
&=
\antisymmetric{\hat{r}_2 \hat{p}_3 – \hat{r}_3 \hat{p}_2}{\hat{r}_3 \hat{p}_1
– \hat{r}_1 \hat{p}_3} \\
&=
\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_3 \hat{p}_1}
-\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_1 \hat{p}_3}
-\antisymmetric{\hat{r}_3 \hat{p}_2}{\hat{r}_3 \hat{p}_1}
+\antisymmetric{\hat{r}_3 \hat{p}_2}{\hat{r}_1 \hat{p}_3}.
\end{aligned}

The first of these commutators is

\label{eqn:qmLecture13:1660}
\begin{aligned}
\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_3 \hat{p}_1}
&=
{\hat{r}_2 \hat{p}_3}{\hat{r}_3 \hat{p}_1}

{\hat{r}_3 \hat{p}_1}
{\hat{r}_2 \hat{p}_3} \\
&=
\hat{r}_2 \hat{p}_1 \antisymmetric{\hat{p}_3}{\hat{r}_3} \\
&=
-i \Hbar \hat{r}_2 \hat{p}_1.
\end{aligned}

We see that any factors in the commutator don’t have like indexes (i.e. $$\hat{r}_k, \hat{p}_k$$) on both position and momentum terms, can be pulled out of the commutator. This leaves

\label{eqn:qmLecture13:1680}
\begin{aligned}
\antisymmetric{\hat{L}_1}{\hat{L}_2}
&=
\hat{r}_2 \hat{p}_1 \antisymmetric{\hat{p}_3}{\hat{r}_3}
-{\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_1 \hat{p}_3}}
-{\antisymmetric{\hat{r}_3 \hat{p}_2}{\hat{r}_3 \hat{p}_1}}
+\hat{r}_1 \hat{p}_2 \antisymmetric{\hat{r}_3}{\hat{p}_3} \\
&=
i \Hbar \lr{ \hat{r}_1 \hat{p}_2 – \hat{r}_2 \hat{p}_1 } \\
&=
i \Hbar \hat{L}_3.
\end{aligned}

With cyclic permutation this is really enough to consider \ref{eqn:qmLecture13:1620} proven. However, can we do this in the general case with the abstract index expression? The quantity to simplify looks forbidding

\label{eqn:qmLecture13:1700}
\antisymmetric{\hat{L}_i}{\hat{L}_j}
=
\epsilon_{i a b }
\epsilon_{j s t }
\antisymmetric{ \hat{r}_a \hat{p}_b }{ \hat{r}_s \hat{p}_t }

Because there are no repeated indexes, this doesn’t submit to any of the normal reduction identities. Note however, since we only care about the $$i \ne j$$ case, that one of the indexes $$a, b$$ must be $$j$$ for this quantity to be non-zero. Therefore (for $$i \ne j$$)

\label{eqn:qmLecture13:1720}
\begin{aligned}
\antisymmetric{\hat{L}_i}{\hat{L}_j}
&=
\epsilon_{i j b }
\epsilon_{j s t }
\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_s \hat{p}_t }
+
\epsilon_{i a j }
\epsilon_{j s t }
\antisymmetric{ \hat{r}_a \hat{p}_j }{ \hat{r}_s \hat{p}_t } \\
&=
\epsilon_{i j b }
\epsilon_{j s t }
\lr{
\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_s \hat{p}_t }

\antisymmetric{ \hat{r}_b \hat{p}_j }{ \hat{r}_s \hat{p}_t }
} \\
&=
-\delta^{s t}_{[i b]}
\antisymmetric{ \hat{r}_j \hat{p}_b – \hat{r}_b \hat{p}_j }{ \hat{r}_s
\hat{p}_t } \\
&=
\antisymmetric{ \hat{r}_j \hat{p}_b – \hat{r}_b \hat{p}_j }{ \hat{r}_b
\hat{p}_i – \hat{r}_i \hat{p}_b } \\
&=
\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_b \hat{p}_i }
– {\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_i \hat{p}_b }}
– {\antisymmetric{ \hat{r}_b \hat{p}_j }{ \hat{r}_b \hat{p}_i }}
+ \antisymmetric{ \hat{r}_b \hat{p}_j }{ \hat{r}_i \hat{p}_b } \\
&=
\hat{r}_j \hat{p}_i \antisymmetric{ \hat{p}_b }{ \hat{r}_b }
+ \hat{r}_i \hat{p}_j \antisymmetric{ \hat{r}_b }{ \hat{p}_b } \\
&=
i \Hbar \lr{ \hat{r}_i \hat{p}_j – \hat{r}_j \hat{p}_i } \\
&=
i \Hbar \epsilon_{i j k} \hat{r}_i \hat{p}_j .
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Plane wave ground state expectation for SHO

Problem [1] 2.18 is, for a 1D SHO, show that

\label{eqn:exponentialExpectationGroundState:20}
\bra{0} e^{i k x} \ket{0} = \exp\lr{ -k^2 \bra{0} x^2 \ket{0}/2 }.

Despite the simple appearance of this problem, I found this quite involved to show. To do so, start with a series expansion of the expectation

\label{eqn:exponentialExpectationGroundState:40}
\bra{0} e^{i k x} \ket{0}
=
\sum_{m=0}^\infty \frac{(i k)^m}{m!} \bra{0} x^m \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:60}
X = \lr{ a + a^\dagger },

so that

\label{eqn:exponentialExpectationGroundState:80}
x
= \sqrt{\frac{\Hbar}{2 \omega m}} X
= \frac{x_0}{\sqrt{2}} X.

Consider the first few values of $$\bra{0} X^n \ket{0}$$

\label{eqn:exponentialExpectationGroundState:100}
\begin{aligned}
\bra{0} X \ket{0}
&=
\bra{0} \lr{ a + a^\dagger } \ket{0} \\
&=
\braket{0}{1} \\
&=
0,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:120}
\begin{aligned}
\bra{0} X^2 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^2 \ket{0} \\
&=
\braket{1}{1} \\
&=
1,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:140}
\begin{aligned}
\bra{0} X^3 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^3 \ket{0} \\
&=
\bra{1} \lr{ \sqrt{2} \ket{2} + \ket{0} } \\
&=
0.
\end{aligned}

Whenever the power $$n$$ in $$X^n$$ is even, the braket can be split into a bra that has only contributions from odd eigenstates and a ket with even eigenstates. We conclude that $$\bra{0} X^n \ket{0} = 0$$ when $$n$$ is odd.

Noting that $$\bra{0} x^2 \ket{0} = \ifrac{x_0^2}{2}$$, this leaves

\label{eqn:exponentialExpectationGroundState:160}
\begin{aligned}
\bra{0} e^{i k x} \ket{0}
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \bra{0} x^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \lr{ \frac{x_0^2}{2} }^m \bra{0} X^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{1}{(2 m)!} \lr{ -k^2 \bra{0} x^2 \ket{0} }^m \bra{0} X^{2m} \ket{0}.
\end{aligned}

This problem is now reduced to showing that

\label{eqn:exponentialExpectationGroundState:180}
\frac{1}{(2 m)!} \bra{0} X^{2m} \ket{0} = \inv{m! 2^m},

or

\label{eqn:exponentialExpectationGroundState:200}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&= \frac{(2m)!}{m! 2^m} \\
&= \frac{ (2m)(2m-1)(2m-2) \cdots (2)(1) }{2^m m!} \\
&= \frac{ 2^m (m)(2m-1)(m-1)(2m-3)(m-2) \cdots (2)(3)(1)(1) }{2^m m!} \\
&= (2m-1)!!,
\end{aligned}

where $$n!! = n(n-2)(n-4)\cdots$$.

It looks like $$\bra{0} X^{2m} \ket{0}$$ can be expanded by inserting an identity operator and proceeding recursively, like

\label{eqn:exponentialExpectationGroundState:220}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&=
\bra{0} X^2 \lr{ \sum_{n=0}^\infty \ket{n}\bra{n} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^2 \lr{ \ket{0}\bra{0} + \ket{2}\bra{2} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^{2m-2} \ket{0} + \bra{0} X^2 \ket{2} \bra{2} X^{2m-2} \ket{0}.
\end{aligned}

This has made use of the observation that $$\bra{0} X^2 \ket{n} = 0$$ for all $$n \ne 0,2$$. The remaining term includes the factor

\label{eqn:exponentialExpectationGroundState:240}
\begin{aligned}
\bra{0} X^2 \ket{2}
&=
\bra{0} \lr{a + a^\dagger}^2 \ket{2} \\
&=
\lr{ \bra{0} + \sqrt{2} \bra{2} } \ket{2} \\
&=
\sqrt{2},
\end{aligned}

Since $$\sqrt{2} \ket{2} = \lr{a^\dagger}^2 \ket{0}$$, the expectation of interest can be written

\label{eqn:exponentialExpectationGroundState:260}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + \bra{0} a^2 X^{2m-2} \ket{0}.

How do we expand the second term. Let’s look at how $$a$$ and $$X$$ commute

\label{eqn:exponentialExpectationGroundState:280}
\begin{aligned}
a X
&=
\antisymmetric{a}{X} + X a \\
&=
\antisymmetric{a}{a + a^\dagger} + X a \\
&=
\antisymmetric{a}{a^\dagger} + X a \\
&=
1 + X a,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:300}
\begin{aligned}
a^2 X
&=
a \lr{ a X } \\
&=
a \lr{ 1 + X a } \\
&=
a + a X a \\
&=
a + \lr{ 1 + X a } a \\
&=
2 a + X a^2.
\end{aligned}

Proceeding to expand $$a^2 X^n$$ we find
\label{eqn:exponentialExpectationGroundState:320}
\begin{aligned}
a^2 X^3 &= 6 X + 6 X^2 a + X^3 a^2 \\
a^2 X^4 &= 12 X^2 + 8 X^3 a + X^4 a^2 \\
a^2 X^5 &= 20 X^3 + 10 X^4 a + X^5 a^2 \\
a^2 X^6 &= 30 X^4 + 12 X^5 a + X^6 a^2.
\end{aligned}

It appears that we have
\label{eqn:exponentialExpectationGroundState:340}
\antisymmetric{a^2 X^n}{X^n a^2} = \beta_n X^{n-2} + 2 n X^{n-1} a,

where

\label{eqn:exponentialExpectationGroundState:360}
\beta_n = \beta_{n-1} + 2 (n-1),

and $$\beta_2 = 2$$. Some goofing around shows that $$\beta_n = n(n-1)$$, so the induction hypothesis is

\label{eqn:exponentialExpectationGroundState:380}
\antisymmetric{a^2 X^n}{X^n a^2} = n(n-1) X^{n-2} + 2 n X^{n-1} a.

Let’s check the induction
\label{eqn:exponentialExpectationGroundState:400}
\begin{aligned}
a^2 X^{n+1}
&=
a^2 X^{n} X \\
&=
\lr{ n(n-1) X^{n-2} + 2 n X^{n-1} a + X^n a^2 } X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} a X + X^n a^2 X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} \lr{ 1 + X a } + X^n \lr{ 2 a + X a^2 } \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} + 2 n X^{n} a
+ 2 X^n a
+ X^{n+1} a^2 \\
&=
X^{n+1} a^2 + (2 + 2 n) X^{n} a + \lr{ 2 n + n(n-1) } X^{n-1} \\
&=
X^{n+1} a^2 + 2(n + 1) X^{n} a + (n+1) n X^{n-1},
\end{aligned}

which concludes the induction, giving

\label{eqn:exponentialExpectationGroundState:420}
\bra{ 0 } a^2 X^{n} \ket{0 } = n(n-1) \bra{0} X^{n-2} \ket{0},

and

\label{eqn:exponentialExpectationGroundState:440}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + (2m-2)(2m-3) \bra{0} X^{2m-4} \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:460}
\sigma_{n} = \bra{0} X^n \ket{0},

so that the recurrence relation, for $$2n \ge 4$$ is

\label{eqn:exponentialExpectationGroundState:480}
\sigma_{2n} = \sigma_{2n -2} + (2n-2)(2n-3) \sigma_{2n -4}

We want to show that this simplifies to

\label{eqn:exponentialExpectationGroundState:500}
\sigma_{2n} = (2n-1)!!

The first values are

\label{eqn:exponentialExpectationGroundState:540}
\sigma_0 = \bra{0} X^0 \ket{0} = 1

\label{eqn:exponentialExpectationGroundState:560}
\sigma_2 = \bra{0} X^2 \ket{0} = 1

which gives us the right result for the first term in the induction

\label{eqn:exponentialExpectationGroundState:580}
\begin{aligned}
\sigma_4
&= \sigma_2 + 2 \times 1 \times \sigma_0 \\
&= 1 + 2 \\
&= 3!!
\end{aligned}

For the general induction term, consider

\label{eqn:exponentialExpectationGroundState:600}
\begin{aligned}
\sigma_{2n + 2}
&= \sigma_{2n} + 2 n (2n – 1) \sigma_{2n -2} \\
&= (2n-1)!! + 2n ( 2n – 1) (2n -3)!! \\
&= (2n + 1) (2n -1)!! \\
&= (2n + 1)!!,
\end{aligned}

which completes the final induction. That was also the last thing required to complete the proof, so we are done!

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## (a)

For a 1D SHO, compute $$\bra{m} x \ket{n}, \bra{m} x^2 \ket{n}, \bra{m} p \ket{n}, \bra{m} p^2 \ket{n}$$ and $$\bra{m} \symmetric{x}{p} \ket{n}$$.

## (b)

Verify the virial theorem is satisfied for energy eigenstates.

## (a)

Using

\label{eqn:shoExpectations:20}
\begin{aligned}
x &= \frac{x_0}{\sqrt{2}} \lr{ a + a^\dagger } \\
p &= \frac{i\Hbar}{x_0 \sqrt{2}} \lr{ a^\dagger – a} \\
a(t) &= a(0) e^{-i \omega t} \\
a(0) \ket{n} &= \sqrt{n} \ket{n-1} \\
a^\dagger(0) \ket{n} &= \sqrt{n+1} \ket{n+1} \\
x_0^2 &= \frac{\Hbar}{\omega m},
\end{aligned}

we have

\label{eqn:shoExpectations:40}
\begin{aligned}
\bra{m} x \ket{n}
&=
\frac{x_0}{\sqrt{2}} \bra{m} \lr{ a + a^\dagger } \ket{n} \\
&=
\frac{x_0}{\sqrt{2}} \bra{m}
\lr{
e^{-i \omega t} \sqrt{n} \ket{n-1}
+
e^{i \omega t} \sqrt{n+1} \ket{n+1}
} \\
&=
\frac{x_0}{\sqrt{2}} \lr{
\delta_{m, n-1} e^{-i \omega t} \sqrt{n}
+
\delta_{m, n+1} e^{i \omega t} \sqrt{n+1}
},
\end{aligned}

\label{eqn:shoExpectations:60}
\begin{aligned}
\bra{m} x^2 \ket{n}
&=
\frac{x_0^2}{2} \bra{m} \lr{ a + a^\dagger }^2 \ket{n} \\
&=
\frac{x_0^2}{2}
\lr{
e^{i \omega t} \sqrt{m} \bra{m-1}
+
e^{-i \omega t} \sqrt{m+1} \bra{m+1}
}
\lr{
e^{-i \omega t} \sqrt{n} \ket{n-1}
+
e^{i \omega t} \sqrt{n+1} \ket{n+1}
} \\
&=
\frac{x_0^2}{2}
\lr{
\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}
+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}
+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}
+\delta_{m-1,n-1} \sqrt{m n}
},
\end{aligned}

\label{eqn:shoExpectations:80}
\begin{aligned}
\bra{m} p \ket{n}
&=
\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{ a^\dagger – a} \ket{n} \\
&=
\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{
e^{i \omega t} \sqrt{n+1} \ket{n+1}

e^{-i \omega t} \sqrt{n} \ket{n-1}
} \\
&=
\frac{i\Hbar}{\sqrt{2} x_0} \lr{
\delta_{m,n+1} e^{i \omega t} \sqrt{n+1}

\delta_{m,n-1} e^{-i \omega t} \sqrt{n}
},
\end{aligned}

\label{eqn:shoExpectations:100}
\begin{aligned}
\bra{m} p^2 \ket{n}
&=
\frac{\Hbar^2}{2 x_0^2} \ket{m} \lr{ a – a^\dagger } \lr{ a^\dagger – a}
\ket{n} \\
&=
\frac{\Hbar^2}{2 x_0^2}
\lr{
-e^{-i \omega t} \sqrt{m+1} \bra{m+1}
+
e^{i \omega t} \sqrt{m} \bra{m-1}
}
\lr{
e^{i \omega t} \sqrt{n+1} \ket{n+1}

e^{-i \omega t} \sqrt{n} \ket{n-1}
} \\
&=
\frac{\Hbar^2}{2 x_0^2}
\lr{
\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}
+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}
+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}
+\delta_{m-1,n-1} \sqrt{m n}
}.
\end{aligned}

For the anticommutator $$\symmetric{x}{p}$$, we have

\label{eqn:shoExpectations:120}
\begin{aligned}
\symmetric{x}{p}
&=
\frac{i\Hbar}{2}
\lr{
\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} } \lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }

\lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }
\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} }
} \\
&=
\frac{i\Hbar}{2}
\lr{
– a^2 e^{- 2 i \omega t}
+ (a^\dagger)^2 e^{ 2 i \omega t}
+ a a^\dagger
– a^\dagger a
+ a^2 e^{- 2 i \omega t}
– (a^\dagger)^2 e^{ 2 i \omega t}
– a^\dagger a
+ a a^\dagger
} \\
&=
i\Hbar
\lr{
a a^\dagger – a^\dagger a
},
\end{aligned}

so

\label{eqn:shoExpectations:140}
\begin{aligned}
\bra{m} \symmetric{x}{p} \ket{n}
&=
i\Hbar
\bra{m}
\lr{
a a^\dagger – a^\dagger a
}
\ket{n} \\
&=
i\Hbar
\bra{m}
\lr{
\sqrt{(n+1)^2}\ket{n}
-\sqrt{n^2}\ket{n}
} \\
&=
i\Hbar
\bra{m}
\lr{
2 n + 1
}
\ket{n}.
\end{aligned}

## (b)

For the SHO, the virial theorem requires $$\expectation{p^2/m} = \expectation{m \omega x^2}$$. That momentum expectation with respect to the eigenstate $$\ket{n}$$ is

\label{eqn:shoExpectations:160}
\begin{aligned}
\expectation{p^2/m}
&=
\frac{\Hbar^2}{2 x_0^2 m}
\lr{
\sqrt{(n+1)(n+1)}
+
\sqrt{n n}
} \\
&=
\frac{\Hbar^2 m \omega}{2 \Hbar m} \lr{ 2 n + 1 } \\
&=
\Hbar \omega \lr{ n + \inv{2} }.
\end{aligned}

For the position expectation we’ve got

\label{eqn:shoExpectations:180}
\begin{aligned}
\expectation{m \omega x^2}
&=
\frac{m \omega^2 x_0^2}{2}
\lr{
\sqrt{(n+1)(n+1)}
+ \sqrt{n n}
} \\
&=
\frac{m \omega^2 \Hbar}{2 m \omega}
\lr{
\sqrt{(n+1)(n+1)}
+ \sqrt{n n}
} \\
&=
\frac{\omega \Hbar}{2 }
\lr{ 2 n + 1 } \\
&=
\omega \Hbar
\lr{ n + \inv{2} }.
\end{aligned}

This shows that the virial theorem holds for the SHO Hamiltonian for eigenstates.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## PHY1520H Graduate Quantum Mechanics. Lecture 5: time evolution of coherent states, and charged particles in a magnetic field. Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering \textchapref{{1}} [1] content.

### Coherent states (cont.)

A coherent state for the SHO $$H = \lr{ N + \inv{2} } \Hbar \omega$$ was given by

\label{eqn:qmLecture5:20}
a \ket{z} = z \ket{z},

where we showed that

\label{eqn:qmLecture5:40}
\ket{z} = c_0 e^{ z a^\dagger } \ket{0}.

In the Heisenberg picture we found

\label{eqn:qmLecture5:60}
\begin{aligned}
a_{\textrm{H}}(t) &= e^{i H t/\Hbar} a e^{-i H t/\Hbar} = a e^{-i\omega t} \\
a_{\textrm{H}}^\dagger(t) &= e^{i H t/\Hbar} a^\dagger e^{-i H t/\Hbar} = a^\dagger e^{i\omega t}.
\end{aligned}

Recall that the position and momentum representation of the ladder operators was

\label{eqn:qmLecture5:80}
\begin{aligned}
a &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} + i \hat{p} \sqrt{\inv{m \Hbar \omega}} } \\
a^\dagger &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} – i \hat{p} \sqrt{\inv{m \Hbar \omega}} },
\end{aligned}

or equivalently
\label{eqn:qmLecture5:100}
\begin{aligned}
\hat{x} &= \lr{ a + a^\dagger } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
\hat{p} &= i \lr{ a^\dagger – a } \sqrt{\frac{m \Hbar \omega}{2}}.
\end{aligned}

Given this we can compute expectation value of position operator

\label{eqn:qmLecture5:120}
\begin{aligned}
\bra{z} \hat{x} \ket{z}
&=
\sqrt{\frac{\Hbar}{ 2 m \omega}}
\bra{z}
\lr{ a + a^\dagger }
\ket{z} \\
&=
\lr{ z + z^\conj } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
&=
2 \textrm{Re} z \sqrt{\frac{\Hbar}{ 2 m \omega}} .
\end{aligned}

Similarly

\label{eqn:qmLecture5:140}
\begin{aligned}
\bra{z} \hat{p} \ket{z}
&=
i \sqrt{\frac{m \Hbar \omega}{2}}
\bra{z}
\lr{ a^\dagger – a }
\ket{z} \\
&=
\sqrt{\frac{m \Hbar \omega}{2}}
2 \textrm{Im} z.
\end{aligned}

How about the expectation of the Heisenberg position operator? That is

\label{eqn:qmLecture5:160}
\begin{aligned}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z}
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \bra{z} \lr{ a + a^\dagger } \ket{z} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ z e^{-i \omega t} + z^\conj e^{i \omega t}} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \lr{z + z^\conj} \cos( \omega t ) -i \lr{ z – z^\conj } \sin( \omega t) } \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \expectation{x(0)} \sqrt{ \frac{2 m \omega}{\Hbar}} \cos( \omega t ) -i \expectation{p(0)} i \sqrt{\frac{2 m \omega}{\Hbar} } \sin( \omega t) } \\
&=
\expectation{x(0)} \cos( \omega t ) + \frac{\expectation{p(0)}}{m \omega} \sin( \omega t) .
\end{aligned}

We find that the average of the Heisenberg position operator evolves in time in exactly the same fashion as position in the classical Harmonic oscillator. This phase space like trajectory is sketched in fig. 1.

fig. 1. phase space like trajectory

In the text it is shown that we have the same structure for the Heisenberg operator itself, before taking expectations

\label{eqn:qmLecture5:220}
\hat{x}_{\textrm{H}}(t)
=
{x(0)} \cos( \omega t ) + \frac{{p(0)}}{m \omega} \sin( \omega t).

Where the coherent states become useful is that we will see that the second moments of position and momentum are not time dependent with respect to the coherent states. Such states remain localized.

### Uncertainty

First note that using the commutator relationship we have

\label{eqn:qmLecture5:180}
\begin{aligned}
\bra{z} a a^\dagger \ket{z}
&=
\bra{z} \lr{ \antisymmetric{a}{a^\dagger} + a^\dagger a } \ket{z} \\
&=
\bra{z} \lr{ 1 + a^\dagger a } \ket{z}.
\end{aligned}

For the second moment we have

\label{eqn:qmLecture5:200}
\begin{aligned}
\bra{z} \hat{x}^2 \ket{z}
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{a + a^\dagger } \lr{a + a^\dagger } \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + a a^\dagger + a^\dagger a
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + 2 a^\dagger a + 1
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z^2 + {(z^\conj)}^2 + 2 z^\conj z + 1} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z + z^\conj }^2
+
\frac{\Hbar}{ 2 m \omega}.
\end{aligned}

We find

\label{eqn:qmLecture5:240}
\sigma_x^2 = \frac{\Hbar}{ 2 m \omega},

and

\label{eqn:qmLecture5:260}
\sigma_p^2 = \frac{m \Hbar \omega}{2}

so

\label{eqn:qmLecture5:280}
\sigma_x^2 \sigma_p^2 = \frac{\Hbar^2}{4},

or

\label{eqn:qmLecture5:300}
\sigma_x \sigma_p = \frac{\Hbar}{2}.

This is the minimum uncertainty.

### Quantum Field theory

In Quantum Field theory the ideas of isolated oscillators is used to model particle creation. The lowest energy state (a no particle, vacuum state) is given the lowest energy level, with each additional quantum level modeling a new particle creation state as sketched in fig. 2.

fig. 2. QFT energy levels

We have to imagine many oscillators, each with a distinct vacuum energy $$\sim \Bk^2$$ . The Harmonic oscillator can be used to model the creation of particles with $$\Hbar \omega$$ energy differences from that “vacuum energy”.

### Charged particle in a magnetic field

In the classical case ( with SI units or $$c = 1$$ ) we have

\label{eqn:qmLecture5:320}
\BF = q \BE + q \Bv \cross \BB.

Alternately, we can look at the Hamiltonian view of the system, written in terms of potentials

\label{eqn:qmLecture5:340}

\label{eqn:qmLecture5:360}
\BE = – \spacegrad \phi – \PD{t}{\BA}.

Note that the curl form for the magnetic field implies one of the required Maxwell’s equations $$\spacegrad \cdot \BB = 0$$.

Ignoring time dependence of the potentials, the Hamiltonian can be expressed as

\label{eqn:qmLecture5:380}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.

In this Hamiltonian the vector $$\Bp$$ is called the canonical momentum, the momentum conjugate to position in phase space.

It is left as an exercise to show that the Lorentz force equation results from application of the Hamiltonian equations of motion, and that the velocity is given by $$\Bv = (\Bp – q \BA)/m$$.

For quantum mechanics, we use the same Hamiltonian, but promote our position, momentum and potentials to operators.

\label{eqn:qmLecture5:400}
\hat{H} = \inv{2 m} \lr{ \hat{\Bp} – q \hat{\BA}(\Br, t) }^2 + q \hat{\phi}(\Br, t).

### Gauge invariance

Can we say anything about this before looking at the question of a particle in a magnetic field?

Recall that the we can make a gauge transformation of the form

\label{eqn:qmLecture5:420a}
\label{eqn:qmLecture5:420}
\BA \rightarrow \BA + \spacegrad \chi

\label{eqn:qmLecture5:440}
\phi \rightarrow \phi – \PD{t}{\chi}

Does this notion of gauge invariance also carry over to the Quantum Hamiltonian. After gauge transformation we have

\label{eqn:qmLecture5:460}
\hat{H}’
= \inv{2 m} \lr{ \hat{\Bp} – q \BA – q \spacegrad \chi }^2 + q \lr{ \phi – \PD{t}{\chi} }

Now we are in a mess, since this function $$\chi$$ can make the Hamiltonian horribly complicated. We don’t see how gauge invariance can easily be applied to the quantum problem. Next time we will introduce a transformation that resolves some of this mess.

## Question: Lorentz force from classical electrodynamic Hamiltonian

Given the classical Hamiltonian

\label{eqn:qmLecture5:381}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.

apply the Hamiltonian equations of motion

\label{eqn:qmLecture5:480}
\begin{aligned}
\ddt{\Bp} &= – \PD{\Bq}{H} \\
\ddt{\Bq} &= \PD{\Bp}{H},
\end{aligned}

to show that this is the Hamiltonian that describes the Lorentz force equation, and to find the velocity in terms of the canonical momentum and vector potential.

The particle velocity follows easily

\label{eqn:qmLecture5:500}
\begin{aligned}
\Bv
&= \ddt{\Br} \\
&= \PD{\Bp}{H} \\
&= \inv{m} \lr{ \Bp – a \BA }.
\end{aligned}

For the Lorentz force we can proceed in the coordinate representation

\label{eqn:qmLecture5:520}
\begin{aligned}
\ddt{p_k}
&= – \PD{x_k}{H} \\
&= – \frac{2}{2m} \lr{ p_m – q A_m } \PD{x_k}{}\lr{ p_m – q A_m } – q \PD{x_k}{\phi} \\
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
\end{aligned}

We also have

\label{eqn:qmLecture5:540}
\begin{aligned}
\ddt{p_k}
&=
\ddt{} \lr{m x_k + q A_k } \\
&=
m \frac{d^2 x_k}{dt^2} + q \PD{x_m}{A_k} \frac{d x_m}{dt} + q \PD{t}{A_k}.
\end{aligned}

Putting these together we’ve got

\label{eqn:qmLecture5:560}
\begin{aligned}
m \frac{d^2 x_k}{dt^2}
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
– q \PD{x_m}{A_k} \frac{d x_m}{dt} – q \PD{t}{A_k} \\
&=
q v_m \lr{ \PD{x_k}{A_m} – \PD{x_m}{A_k} } + q E_k \\
&=
q v_m \epsilon_{k m s} B_s + q E_k,
\end{aligned}

or

\label{eqn:qmLecture5:580}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&=
q \Be_k v_m \epsilon_{k m s} B_s + q E_k \\
&= q \Bv \cross \BB + q \BE.
\end{aligned}

## Question: Show gauge invariance of the magnetic and electric fields

After the gauge transformation of \ref{eqn:qmLecture5:420} show that the electric and magnetic fields are unaltered.

For the magnetic field the transformed field is

\label{eqn:qmLecture5:600}
\begin{aligned}
\BB’
&= \BB.
\end{aligned}

\label{eqn:qmLecture5:620}
\begin{aligned}
\BE’
&=
– \PD{t}{\BA’} – \spacegrad \phi’ \\
&=
– \PD{t}{}\lr{\BA + \spacegrad \chi} – \spacegrad \lr{ \phi – \PD{t}{\chi}} \\
&=
– \PD{t}{\BA} – \spacegrad \phi \\
&=
\BE.
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## PHY1520H Graduate Quantum Mechanics. Lecture 4: Quantum Harmonic oscillator and coherent states. Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough. This lecture reviewed a lot of quantum harmonic oscillator theory, and wouldn’t make sense without having seen raising and lowering operators (ladder operators), number operators, and the like.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 2 content.

### Classical Harmonic Oscillator

Recall the classical Harmonic oscillator equations in their Hamiltonian form

\label{eqn:qmLecture4:40}
\ddt{x} = \frac{p}{m}

\label{eqn:qmLecture4:60}
\ddt{p} = -k x.

With

\label{eqn:qmLecture4:140}
\begin{aligned}
x(t = 0) &= x_0 \\
p(t = 0) &= p_0 \\
k &= m \omega^2,
\end{aligned}

the solutions are ellipses in phase space

\label{eqn:qmLecture4:100}
x(t) = x_0 \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t)

\label{eqn:qmLecture4:120}
p(t) = p_0 \cos(\omega t) – m \omega x_0 \sin(\omega t).

After a suitable scaling of the variables, these elliptical orbits can be transformed into circular trajectories.

### Quantum Harmonic Oscillator

\label{eqn:qmLecture4:160}
\hat{H} = \frac{\hat{p}^2}{2 m} + \inv{2} k \hat{x}^2

Set

\label{eqn:qmLecture4:200}
\hat{X} = \sqrt{\frac{m \omega}{\Hbar}} \hat{x}

\label{eqn:qmLecture4:220}
\hat{P} = \sqrt{\inv{m \omega \Hbar}} \hat{p}

The commutators after this change of variables goes from

\label{eqn:qmLecture4:240}
\antisymmetric{ \hat{x}}{\hat{p}} = i \Hbar,

to
\label{eqn:qmLecture4:260}
\antisymmetric{ \hat{X}}{\hat{P}} = i.

The Hamiltonian takes the form

\label{eqn:qmLecture4:280}
\begin{aligned}
\hat{H}
&= \frac{\Hbar \omega}{2} \lr{ \hat{X}^2 + \hat{P}^2 } \\
&= \Hbar \omega \lr{ \lr{ \frac{\hat{X} -i \hat{P}}{\sqrt{2}} } \lr{ \frac{\hat{X} +i \hat{P}}{\sqrt{2}}} + \inv{2} }.
\end{aligned}

Define ladder operators (raising and lowering operators respectively)

\label{eqn:qmLecture4:320}
\hat{a}^\dagger = \frac{\hat{X} -i \hat{P}}{\sqrt{2}}

\label{eqn:qmLecture4:340}
\hat{a} = \frac{\hat{X} +i \hat{P}}{\sqrt{2}}

so

\label{eqn:qmLecture4:360}
\hat{H} = \Hbar \omega \lr{ \hat{a}^\dagger \hat{a} + \inv{2} }.

We can show

\label{eqn:qmLecture4:380}
\antisymmetric{\hat{a}}{\hat{a}^\dagger} = 1,

and

\label{eqn:qmLecture4:400}
N \ket{n} \equiv \hat{a}^\dagger a = n \ket{n},

where $$n \ge 0$$ is an integer. Recall that

\label{eqn:qmLecture4:420}
\hat{a} \ket{0} = 0,

and

\label{eqn:qmLecture4:440}
\bra{X} X + i P \ket{0} = 0.

With

\label{eqn:qmLecture4:460}
\braket{x}{0} = \Psi_0(x),

we can show

\label{eqn:qmLecture4:480}
\inv{\sqrt{2}} \lr{ X + \PD{X}{} } \Psi_0(X) = 0.

Also recall that

\label{eqn:qmLecture4:520}
\hat{a} \ket{n} = \sqrt{n} \ket{n-1}

\label{eqn:qmLecture4:540}
\hat{a}^\dagger \ket{n} = \sqrt{n + 1} \ket{n+1}

### Coherent states

Coherent states for the quantum harmonic oscillator are the eigenkets for the creation and annihilation operators

\label{eqn:qmLecture4:580}
\hat{a} \ket{z} = z \ket{z}

\label{eqn:qmLecture4:600}
\hat{a}^\dagger \ket{\tilde{z}} = \tilde{z} \ket{\tilde{z}} ,

where

\label{eqn:qmLecture4:620}
\ket{z} = \sum_{n = 0}^\infty c_n \ket{n},

and $$z$$ is allowed to be a complex number.

Looking for such a state, we compute

\label{eqn:qmLecture4:640}
\begin{aligned}
\hat{a} \ket{z}
&= \sum_{n=1}^\infty c_n \hat{a} \ket{n} \\
&= \sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1}
\end{aligned}

compare this to

\label{eqn:qmLecture4:660}
\begin{aligned}
z \ket{z}
&=
z \sum_{n=0}^\infty c_n \ket{n} \\
&=
\sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1} \\
&=
\sum_{n=0}^\infty c_{n+1} \sqrt{n+1} \ket{n},
\end{aligned}

so

\label{eqn:qmLecture4:680}
c_{n+1} \sqrt{n+1} = z c_n

This gives

\label{eqn:qmLecture4:700}
c_{n+1} = \frac{z c_n}{\sqrt{n+1}}

\label{eqn:qmLecture4:720}
\begin{aligned}
c_1 &= c_0 z \\
c_2 &= \frac{z c_1}{\sqrt{2}} = \frac{z^2 c_0}{\sqrt{2}} \\
\vdots &
\end{aligned}

or

\label{eqn:qmLecture4:740}
c_n = \frac{z^n}{\sqrt{n!}}.

So the desired state is

\label{eqn:qmLecture4:760}
\ket{z} = c_0 \sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}} \ket{n}.

Also recall that

\label{eqn:qmLecture4:780}
\ket{n} = \frac{\lr{ \hat{a}^\dagger }^n}{\sqrt{n!}} \ket{0},

which gives

\label{eqn:qmLecture4:800}
\begin{aligned}
\ket{z}
&= c_0 \sum_{n=0}^\infty \frac{\lr{z \hat{a}^\dagger}^n }{n!} \ket{0} \\
&= c_0 e^{z \hat{a}^\dagger} \ket{0}.
\end{aligned}

The normalization is

\label{eqn:qmLecture4:820}
c_0 = e^{-\Abs{z}^2/2}.

While we have $$\braket{n_1}{n_2} = \delta_{n_1, n_2}$$, these $$\ket{z}$$ states are not orthonormal. Figuring out that this overlap

\label{eqn:qmLecture4:840}
\braket{z_1}{z_2} \ne 0,

will be left for homework.

### Dynamics

We don’t know much about these coherent states. For example does a coherent state at time zero evolve to a coherent state?

\label{eqn:qmLecture4:860}
\ket{z} \stackrel{?}{\rightarrow} \ket{z(t)}

It turns out that these questions are best tackled in the Heisenberg picture, considering

\label{eqn:qmLecture4:880}
e^{-i \hat{H} t/\Hbar } \ket{z}.

For example, what is the average of the position operator

\label{eqn:qmLecture4:900}
\bra{z} e^{i \hat{H} t/\Hbar } \hat{x} e^{-i \hat{H} t/\Hbar } \ket{z}
=
\sum_{n, n’ = 0}^\infty
\bra{n} c_n^\conj e^{i E_n t/\Hbar}
\lr{ a + a^\dagger} \sqrt{ \frac{\Hbar}{m \omega} }
c_{n’} e^{i E_{n’} t/\Hbar}
\ket{n}.

This is very messy to attempt. Instead if we know how the operator evolves we can calculate

\label{eqn:qmLecture4:920}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},

that is

\label{eqn:qmLecture4:940}
\expectation{\hat{x}}(t) = \bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},

and for momentum

\label{eqn:qmLecture4:960}
\expectation{\hat{p}}(t) = \bra{z} \hat{p}_{\textrm{H}}(t) \ket{z}.

The question to ask is what are the expansions of

\label{eqn:qmLecture4:1000}
\hat{a}_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar}.

\label{eqn:qmLecture4:1020}
\hat{a}^\dagger_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a}^\dagger e^{-i \hat{H} t/\Hbar}.

The question to ask is how do these operators ask on the basis states

\label{eqn:qmLecture4:1040}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) \ket{n}
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar} \ket{n} \\
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i t \omega (n + 1/2)} \ket{n} \\
&=
e^{-i t \omega (n + 1/2)}
e^{i \hat{H} t/\Hbar}
\sqrt{n} \ket{n-1} \\
&=
\sqrt{n}
e^{-i t \omega (n + 1/2)}
e^{i t \omega (n – 1/2)}
\ket{n-1} \\
&=
\sqrt{n} e^{-i \omega t} \ket{n-1} \\
&=
e^{-i \omega t} \ket{n}.
\end{aligned}

So we have found

\label{eqn:qmLecture4:1060}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) &= a e^{-i\omega t} \\
\hat{a}^\dagger_{\textrm{H}}(t) &= a^\dagger e^{i\omega t}
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Quantum SHO ladder operators as a diagonal change of basis for the Heisenberg EOMs

Many authors pull the definitions of the raising and lowering (or ladder) operators out of their butt with no attempt at motivation. This is pointed out nicely in [1] by Eli along with one justification based on factoring the Hamiltonian.

In [2] is a small exception to the usual presentation. In that text, these operators are defined as usual with no motivation. However, after the utility of these operators has been shown, the raising and lowering operators show up in a context that does provide that missing motivation as a side effect.
It doesn’t look like the author was trying to provide a motivation, but it can be interpreted that way.

When seeking the time evolution of Heisenberg-picture position and momentum operators, we will see that those solutions can be trivially expressed using the raising and lowering operators. No special tools nor black magic is required to find the structure of these operators. Unfortunately, we must first switch to both the Heisenberg picture representation of the position and momentum operators, and also employ the Heisenberg equations of motion. Neither of these last two fit into standard narrative of most introductory quantum mechanics treatments. We will also see that these raising and lowering “operators” could also be introduced in classical mechanics, provided we were attempting to solve the SHO system using the Hamiltonian equations of motion.

I’ll outline this route to finding the structure of the ladder operators below. Because these are encountered trying to solve the time evolution problem, I’ll first show a simpler way to solve that problem. Because that simpler method depends a bit on lucky observation and is somewhat unstructured, I’ll then outline a more structured procedure that leads to the ladder operators directly, also providing the solution to the time evolution problem as a side effect.

The starting point is the Heisenberg equations of motion. For a time independent Hamiltonian $$H$$, and a Heisenberg operator $$A^{(H)}$$, those equations are

\label{eqn:harmonicOscDiagonalize:20}
\ddt{A^{(H)}} = \inv{i \Hbar} \antisymmetric{A^{(H)}}{H}.

Here the Heisenberg operator $$A^{(H)}$$ is related to the Schrodinger operator $$A^{(S)}$$ by

\label{eqn:harmonicOscDiagonalize:60}
A^{(H)} = U^\dagger A^{(S)} U,

where $$U$$ is the time evolution operator. For this discussion, we need only know that $$U$$ commutes with $$H$$, and do not need to know the specific structure of that operator. In particular, the Heisenberg equations of motion take the form

\label{eqn:harmonicOscDiagonalize:80}
\begin{aligned}
\ddt{A^{(H)}}
&= \inv{i \Hbar}
\antisymmetric{A^{(H)}}{H} \\
&= \inv{i \Hbar}
\antisymmetric{U^\dagger A^{(S)} U}{H} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} U H
– H U^\dagger A^{(S)} U
} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} H U
– U^\dagger H A^{(S)} U
} \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{A^{(S)}}{H} U.
\end{aligned}

The Hamiltonian for the harmonic oscillator, with Schrodinger-picture position and momentum operators $$x, p$$ is

\label{eqn:harmonicOscDiagonalize:40}
H = \frac{p^2}{2m} + \inv{2} m \omega^2 x^2,

so the equations of motions are

\label{eqn:harmonicOscDiagonalize:100}
\begin{aligned}
\ddt{x^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{\frac{p^2}{2m}} U \\
&= \inv{2 m i \Hbar} U^\dagger \lr{ i \Hbar \PD{p}{p^2} } U \\
&= \inv{m } U^\dagger p U \\
&= \inv{m } p^{(H)},
\end{aligned}

and
\label{eqn:harmonicOscDiagonalize:120}
\begin{aligned}
\ddt{p^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{\inv{2} m \omega^2 x^2 } U \\
&= \frac{m \omega^2}{2 i \Hbar} U^\dagger \lr{ -i \Hbar \PD{x}{x^2} } U \\
&= -m \omega^2 U^\dagger x U \\
&= -m \omega^2 x^{(H)}.
\end{aligned}

In the Heisenberg picture the equations of motion are precisely those of classical Hamiltonian mechanics, except that we are dealing with operators instead of scalars

\label{eqn:harmonicOscDiagonalize:140}
\begin{aligned}
\ddt{p^{(H)}} &= -m \omega^2 x^{(H)} \\
\ddt{x^{(H)}} &= \inv{m } p^{(H)}.
\end{aligned}

In the text the ladder operators are used to simplify the solution of these coupled equations, since they can decouple them. That’s not really required since we can solve them directly in matrix form with little work

\label{eqn:harmonicOscDiagonalize:160}
\ddt{}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix}
=
\begin{bmatrix}
0 & -m \omega^2 \\
\inv{m} & 0
\end{bmatrix}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix},

or, with length scaled variables

\label{eqn:harmonicOscDiagonalize:180}
\begin{aligned}
\ddt{}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}
&=
\begin{bmatrix}
0 & -\omega \\
\omega & 0
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\sigma_y
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}.
\end{aligned}

Writing $$y = \begin{bmatrix} \frac{p^{(H)}}{m \omega} \\ x^{(H)} \end{bmatrix}$$, the solution can then be written immediately as

\label{eqn:harmonicOscDiagonalize:200}
\begin{aligned}
y(t)
&=
\exp\lr{ -i \omega \sigma_y t } y(0) \\
&=
\lr{ \cos \lr{ \omega t } I – i \sigma_y \sin\lr{ \omega t } } y(0) \\
&=
\begin{bmatrix}
\cos\lr{ \omega t } & \sin\lr{ \omega t } \\
-\sin\lr{ \omega t } & \cos\lr{ \omega t }
\end{bmatrix}
y(0),
\end{aligned}

or

\label{eqn:harmonicOscDiagonalize:220}
\begin{aligned}
\frac{p^{(H)}(t)}{m \omega} &= \cos\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \sin\lr{ \omega t } x^{(H)}(0) \\
x^{(H)}(t) &= -\sin\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \cos\lr{ \omega t } x^{(H)}(0).
\end{aligned}

This solution depends on being lucky enough to recognize that the matrix has a Pauli matrix as a factor (which squares to unity, and allows the exponential to be evaluated easily.)

If we hadn’t been that observant, then the first tool we’d have used instead would have been to diagonalize the matrix. For such diagonalization, it’s natural to work in completely dimensionless variables. Such a non-dimensionalisation can be had by defining

\label{eqn:harmonicOscDiagonalize:240}
x_0 = \sqrt{\frac{\Hbar}{m \omega}},

and dividing the working (operator) variables through by those values. Let $$z = \inv{x_0} y$$, and $$\tau = \omega t$$ so that the equations of motion are

\label{eqn:harmonicOscDiagonalize:260}
\frac{dz}{d\tau}
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
z.

This matrix can be diagonalized as

\label{eqn:harmonicOscDiagonalize:280}
A
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
=
V
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
V^{-1},

where

\label{eqn:harmonicOscDiagonalize:300}
V =
\inv{\sqrt{2}}
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}.

The equations of motion can now be written

\label{eqn:harmonicOscDiagonalize:320}
\frac{d}{d\tau} \lr{ V^{-1} z } =
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
\lr{ V^{-1} z }.

This final change of variables $$V^{-1} z$$ decouples the system as desired. Expanding that gives

\label{eqn:harmonicOscDiagonalize:340}
\begin{aligned}
V^{-1} z
&=
\inv{\sqrt{2}}
\begin{bmatrix}
-i & 1 \\
i & 1
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i \frac{p^{(H)}}{m \omega} + x^{(H)} \\
i \frac{p^{(H)}}{m \omega} + x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
a^\dagger \\
a
\end{bmatrix},
\end{aligned}

where
\label{eqn:harmonicOscDiagonalize:n}
\begin{aligned}
a^\dagger &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ -i \frac{p^{(H)}}{m \omega} + x^{(H)} } \\
a &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ i \frac{p^{(H)}}{m \omega} + x^{(H)} }.
\end{aligned}

Lo and behold, we have the standard form of the raising and lowering operators, and can write the system equations as

\label{eqn:harmonicOscDiagonalize:360}
\begin{aligned}
\ddt{a^\dagger} &= i \omega a^\dagger \\
\ddt{a} &= -i \omega a.
\end{aligned}

It is actually a bit fluky that this matched exactly, since we could have chosen eigenvectors that differ by constant phase factors, like

\label{eqn:harmonicOscDiagonalize:380}
V = \inv{\sqrt{2}}
\begin{bmatrix}
i e^{i\phi} & -i e^{i \psi} \\
1 e^{i\phi} & e^{i \psi}
\end{bmatrix},

so

\label{eqn:harmonicOscDiagonalize:341}
\begin{aligned}
V^{-1} z
&=
\frac{e^{-i(\phi + \psi)}}{\sqrt{2}}
\begin{bmatrix}
-i e^{i\psi} & e^{i \psi} \\
i e^{i\phi} & e^{i \phi}
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i e^{i\phi} \frac{p^{(H)}}{m \omega} + e^{i\phi} x^{(H)} \\
i e^{i\psi} \frac{p^{(H)}}{m \omega} + e^{i\psi} x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{i\phi} a^\dagger \\
e^{i\psi} a
\end{bmatrix}.
\end{aligned}

To make the resulting pairs of operators Hermitian conjugates, we’d want to constrain those constant phase factors by setting $$\phi = -\psi$$. If we were only interested in solving the time evolution problem no such additional constraints are required.

The raising and lowering operators are seen to naturally occur when seeking the solution of the Heisenberg equations of motion. This is found using the standard technique of non-dimensionalisation and then seeking a change of basis that diagonalizes the system matrix. Because the Heisenberg equations of motion are identical to the classical Hamiltonian equations of motion in this case, what we call the raising and lowering operators in quantum mechanics could also be utilized in the classical simple harmonic oscillator problem. However, in a classical context we wouldn’t have a justification to call this more than a change of basis.

# References

[1] Eli Lansey. The Quantum Harmonic Oscillator Ladder Operators, 2009. URL http://behindtheguesses.blogspot.ca/2009/03/quantum-harmonic-oscillator-ladder.html. [Online; accessed 18-August-2015].

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics, chapter {Time Development of the Oscillator}. Pearson Higher Ed, 2014.