raising operator

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

September 25, 2018 phy2403 , , , , , , , , , , ,

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Disclaimer

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)
\begin{equation}\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
}.
\end{equation}
This Lagrangian density was related to the action by
\begin{equation}\label{eqn:qftLecture5:40}
S = \int dt L = \int dt d^3 x \LL,
\end{equation}
with momentum canonically conjugate to the field \( \phi \) defined as
\begin{equation}\label{eqn:qftLecture5:60}
\Pi(\Bx, t) = \frac{\delta \LL}{\delta \phidot(\Bx, t) } = \PD{\phidot(\Bx, t)}{\LL}
\end{equation}

The Hamiltonian defined as
\begin{equation}\label{eqn:qftLecture5:80}
H = \int d^3 x \lr{ \Pi(\Bx, t) \phidot(\Bx, t) – \LL },
\end{equation}
led to
\begin{equation}\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 }.
\end{equation}
Like the Lagrangian density, we may introduce a Hamiltonian density \( \mathcal{H} \) as
\begin{equation}\label{eqn:qftLecture5:100}
H = \int d^3 x \mathcal{H}(\Bx, t).
\end{equation}
For our Klein-Gordon system, this is
\begin{equation}\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.
\end{equation}

Canonical Commutation Relations (CCR)

:

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

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

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

This is similar to the QM wave functions
\begin{equation}\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}
\end{equation}

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

QM SHO review

Recall the QM SHO had a Hamiltonian
\begin{equation}\label{eqn:qftLecture5:260}
\hat{H} = \inv{2} \hat{p}^2 + \inv{2} \omega^2 \hat{q}^2,
\end{equation}
where
\begin{equation}\label{eqn:qftLecture5:280}
\antisymmetric{\hat{p}}{\hat{q}} = -i,
\end{equation}
and that
HP time evolution operators \( O \) satisfied
\begin{equation}\label{eqn:qftLecture5:700}
\ddt{\hatO} = i \antisymmetric{\hatH}{\hatO}.
\end{equation}
In particular
\begin{equation}\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}
\end{equation}
and
\begin{equation}\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}
\end{equation}
Applying the time evolution operator twice, we find
\begin{equation}\label{eqn:qftLecture5:340}
\frac{d^2}{dt^2}{\hat{q}}
= \ddt{\hat{p}}
= – \omega^2 \hat{q}.
\end{equation}
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
\begin{equation}\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}
\end{equation}

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
\begin{equation}\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).
\end{equation}
\begin{equation}\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.
\end{equation}
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
\begin{equation}\label{eqn:qftLecture5:720}
\begin{aligned}
\int d^3 y (\spacegrad \phihat(\By))^2 \hat{\Pi}(\Bx)
&=
\int d^3 y
\lr{ \spacegrad \phihat(\By) \cdot \spacegrad \phihat(\By) } \hat{\Pi}(\Bx) \\
&=
\int d^3 y
\spacegrad \phihat(\By) \cdot
\lr{ \spacegrad (\phihat(\By) \hat{\Pi}(\Bx)) } \\
&=
\int d^3 y
\spacegrad \phihat(\By) \cdot
\lr{ \spacegrad (\hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By)) } \\
&=
\int d^3 y
\Biglr{
\spacegrad \lr{ \phihat(\By) \hat{\Pi}(\Bx) } \cdot \spacegrad \phihat(\By)
+ i
\spacegrad \phihat(\By) \cdot \spacegrad \delta^3(\Bx – \By)
} \\
&=
\int d^3 y
\Biglr{
\spacegrad \lr{ \hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By) } \cdot \spacegrad \phihat(\By)
+ i
\spacegrad \phihat(\By) \cdot \spacegrad \delta^3(\Bx – \By)
} \\
&=
\int d^3 y
\hat{\Pi}(\Bx)
\lr{
\spacegrad \phihat(\By) \cdot \spacegrad \phihat(\By)
}
+ 2 i
\int d^3 y
\spacegrad \phihat(\By) \cdot \spacegrad \delta^3(\Bx – \By) \\
&=
\int d^3 y
\hat{\Pi}(\Bx) \spacegrad^2 \phihat(\By)
+
2 i
\int d^3 y
\spacegrad \cdot \lr{ \delta^3(\Bx – \By) \spacegrad \phihat(\By) }

2 i
\int d^3 y
\delta^3(\Bx – \By) \spacegrad^2 \phihat(\By) \\
&=
\int d^3 y
\hat{\Pi}(\Bx) \spacegrad^2 \phihat(\By)
+
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)
\ncap \cdot \spacegrad \phihat(\By)

2 i \spacegrad^2 \phihat(\Bx).
\end{aligned}
\end{equation}
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
\begin{equation}\label{eqn:qftLecture5:800}
\begin{aligned}
\int d^3 y \antisymmetric{(\spacegrad \phihat(\By))^2}{\hat{\Pi}(\Bx)}
&=
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)
\ncap \cdot \spacegrad \phihat(\By)

2 i \spacegrad^2 \phihat(\Bx) \\
&=

2 i \spacegrad^2 \phihat(\Bx),
\end{aligned}
\end{equation}
where the boundary integral is presumed to be zero (without enough justification.) All the pieces can now be put back together
\begin{equation}\label{eqn:qftLecture5:820}
\ddt{} \hat{\Pi}(\Bx, t)
=
\spacegrad^2 \phihat(\Bx, t)

m^2 \phihat(\Bx, t)

\lambda \phihat^3(\Bx, t).
\end{equation}

Now, for the \( \phihat \) time evolution, which is much easier
\begin{equation}\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}
\end{equation}
\begin{equation}\label{eqn:qftLecture5:400}
\frac{d^2}{dt^2}{\hat{\phi}}(\Bx, t)
=
\spacegrad^2 \phi
-m^2 \phi – \lambda \phihat^3.
\end{equation}
That is
\begin{equation}\label{eqn:qftLecture5:420}
\ddot{\phihat} – \spacegrad^2 \phihat + m^2 \phihat + \lambda \phihat^3 = 0,
\end{equation}
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
\begin{equation}\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}
\end{equation}

We can discover a representation of the delta function by applying these both in turn
\begin{equation}\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)
\end{equation}
so
\begin{equation}\label{eqn:qftLecture5:500}
\boxed{
\int d^3 x e^{i \BA \cdot \Bx} = (2 \pi)^3 \delta^3(\BA)
}
\end{equation}

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
\begin{equation}\label{eqn:qftLecture5:520}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi(\Bx, t) = 0
\end{equation}

Inserting the transform relation \ref{eqn:qftLecture5:440} we get
\begin{equation}\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,
\end{equation}
or
\begin{equation}\label{eqn:qftLecture5:580}
\boxed{
\ddot{\tilde{\phi}}(\Bp, t) = – \omega_\Bp^2 \,\tilde{\phi}(\Bp, t),
}
\end{equation}
where
\begin{equation}\label{eqn:qftLecture5:560}
\omega_\Bp = \sqrt{ \Bp^2 + m^2 }.
\end{equation}
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
\begin{equation}\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}
}
\end{equation}

Observe that
\begin{equation}\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}
\end{equation}
or
\begin{equation}\label{eqn:qftLecture5:620}
\tilde{\phi}^\dagger(\Bp, t) = \tilde{\phi}(-\Bp, t),
\end{equation}
so \( \phi(\Bp, t) \) has a real representation in terms of \( a_\Bp \).

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

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

Hamiltonian for a scalar field

January 3, 2016 phy2403 , , ,

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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

\begin{equation}\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}
}.
\end{equation}

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
\begin{equation}\label{eqn:scalarFieldHamiltonian:40}
H = \inv{2} \int d^3 x \lr{ \Pi^2 + \lr{ \spacegrad \phi }^2 + \mu^2 \phi^2 }.
\end{equation}

The field derivatives are

\begin{equation}\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}
},
\end{equation}

and

\begin{equation}\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}
}.
\end{equation}

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

\begin{equation}\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}
\end{equation}

For the gradient portion of the Hamiltonian we have

\begin{equation}\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}
\end{equation}

Finally, for the mass term, we have

\begin{equation}\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}
\end{equation}

Now all the pieces can be put back together again

\begin{equation}\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}
\end{equation}

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

References

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

Spin three halves spin interaction

December 15, 2015 phy1520 , , , ,

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Q: [1] pr 3.33

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

\begin{equation}\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
}.
\end{equation}

Show that the interaction energy can be written as

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

Find the energy eigenvalues for such a Hamiltonian.

A:

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

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

The squared spin operators are
\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

This gives
\begin{equation}\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}
\end{equation}

For a static electric field we have

\begin{equation}\label{eqn:spinThreeHalvesNucleus:160}
\spacegrad^2 \phi = -\frac{\rho}{\epsilon_0},
\end{equation}

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

\begin{equation}\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)
},
\end{equation}

so
\begin{equation}\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
}
\end{equation}
\begin{equation}\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 }.
\end{equation}

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 \( \setlr{ \ket{3/2}, \ket{1/2}, \ket{-1/2}, \ket{-3/2} } \)

\begin{equation}\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}
\end{equation}

The energy eigenvalues are found to be

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

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

November 11, 2015 phy1520 , , , ,

[Click here for a PDF of this post with nicer formatting]

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

\begin{equation}\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}
\end{equation}

We were looking for

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

by applying

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

We found

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

Let

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

We want

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

or
\begin{equation}\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}
\end{equation}

We found

\begin{equation}\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}
\end{equation}

so

\begin{equation}\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 }.
\end{equation}

Similarly
\begin{equation}\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 }.
\end{equation}

Constraints

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

If these are satisfied at the equality extreme we have

\begin{equation}\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}
\end{equation}

Rearranging this to solve, we can rewrite the equality as

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

which has solutions at

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

One of the solutions is

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

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

The final constraint is therefore

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

and

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

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

\begin{equation}\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}}},
\end{equation}

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

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

or

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

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

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

so for \( m \in -j, -j + 1, \cdots, +j \)

\begin{equation}\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}
}
\end{equation}

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

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

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

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

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 \( \hat{L}_z = +j \)

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

and for \( \hat{L}_z = -j \)

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

We can make the identifications

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

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

or

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

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.

fig. 3. Number conservation constraint.

Can we make an identification that takes

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

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

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

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

\begin{equation}\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}
\end{equation}

But

\begin{equation}\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}
\end{equation}

and
\begin{equation}\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}
\end{equation}

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

as desired. Similarly

\begin{equation}\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}
\end{equation}

With

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

We can make the identification

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

Another way

With

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

or

\begin{equation}\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},
\end{equation}

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

\begin{equation}\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}
\end{equation}

That gives
\begin{equation}\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}
\end{equation}

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

November 7, 2015 phy1520 , , , ,

[Click here for a PDF of this post with nicer formatting]

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

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

which can alternately be written

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

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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)},
\end{equation}

or

\begin{equation}\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)}.
\end{equation}

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

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

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

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

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

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

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

    but

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

    or

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

    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
    \begin{equation}\label{eqn:qmLecture13:280}
    i \hat{H} = i \Theta^{-1} \hat{H} \Theta,
    \end{equation}

    so

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

    Acting on an energy eigenket, we’ve got

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

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

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

What properties do we expect from \( \Theta \)?

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

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

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

if that’s true, then how about the momentum operator in the position basis
\begin{equation}\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}
\end{equation}

How about the \( x,p \) commutator? For that we have

\begin{equation}\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}
\end{equation}

For the the angular momentum operators

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

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

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

Given a general state
so that if

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

the action of the time reversal operator would be

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

That action is:

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

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

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

produce the desired inversion of sign

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

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

\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

The time reversed actions are

\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

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

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

now try the action

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

This is the action:

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

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

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

Note that acting the time reversal operator twice has the effects

\begin{equation}\label{eqn:qmLecture13:660}
\Theta^2 \ket{\uparrow} = \Theta \ket{\downarrow} = – \ket{\uparrow}
\end{equation}
\begin{equation}\label{eqn:qmLecture13:680}
\Theta^2 \ket{\downarrow} = \Theta (-\ket{\uparrow}) = – \ket{\uparrow}.
\end{equation}

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

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

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

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

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

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

then
\begin{equation}\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}
\end{equation}

but

\begin{equation}\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}
\end{equation}

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

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

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

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

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

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

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

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

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

\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

With this representation, we have

\begin{equation}\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’}.
\end{equation}

\begin{equation}\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’}.
\end{equation}

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

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

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

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

\begin{equation}\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}
\end{equation}

Let’s define
\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\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}
\end{equation}

Explicitly, that is

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

There must be a proportionality of the form

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

The products of the raising and lowering operators are

\begin{equation}\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}
\end{equation}

and
\begin{equation}\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}
\end{equation}

So we must have

\begin{equation}\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}
\end{equation}

and

\begin{equation}\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}
\end{equation}

This puts constraints on \( a, b \), roughly of the form

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

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

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

    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

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

Answer

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

\begin{equation}\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}
\end{equation}

The first of these commutators is

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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 }
\end{equation}

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 \( i \ne j \))

\begin{equation}\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}
\end{equation}

References

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