PHY2403H Quantum Field Theory. Lecture 9: Unbroken and spontaneously broken symmetries, Higgs Lagrangian, scale invariance, Lorentz invariance, angular momentum quantization. Taught by Prof. Erich Poppitz

[Click here for a PDF of this post with nicer formatting (and a Mathematica listing that I didn’t include in this blog post’s latex export)]

DISCLAIMER: Very rough notes from class, with some additional side notes.

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

Last time

We followed a sequence of operations

1. Noether’s theorem
2. $$\rightarrow$$ conserved currents
3. $$\rightarrow$$ charges (classical)
4. $$\rightarrow$$ “correspondence principle”
5. $$\rightarrow \hatQ$$
• Hermitian operators
• “generators of symmetry”
\label{eqn:qftLecture9:20}
\hatU(\alpha) = e^{i \alpha \hatQ}

We found
\label{eqn:qftLecture9:40}
\hatU(\alpha) \phihat \hatU^\dagger(\alpha) = \phihat + i \alpha \antisymmetric{\hatQ}{\phihat} + \cdots

Example: internal symmetries:

(non-spacetime), such as $$O(N)$$ or $$U(1)$$.

In QFT internal symmetries can have different “\underline{modes of realization}”.

[I]

1. “Wigner mode”. These are also called “unbroken symmetries”.
\label{eqn:qftLecture9:60}
\hatQ \ket{0} = 0

i.e. $$\hatU(\alpha) \ket{0} = 0$$.
Ground state invariant. Formally $$:\hatQ:$$ annihilates $$\ket{0}$$.
$$\antisymmetric{\hatQ}{\hatH} = 0$$ implies that all eigenstates are eigenstates of $$\hatQ$$ in $$U(1)$$. Example from HW 1
\label{eqn:qftLecture9:80}
\hatQ = \text{“charge” under $$U(1)$$}.

All states have definite charge, just live in QU.
2. “Nambu-Goldstone mode” (Landau-ginsburg). This is also called a “spontaneously broken symmetry”\footnote{
First encounter example (HWII, $$SU(2) \times SU(2) \rightarrow SU(2)$$). Here a $$U(1)$$ spontaneous broken symmetry.}.
$$H$$ or $$L$$ is invariant under symmetry, but ground state is not.

fig. 1. Mexican hat potential.

fig. 2. Degenerate Mexican hat potential ( v = 0)

Example:
\label{eqn:qftLecture9:100}
\LL = \partial_\mu \phi^\conj \partial^\mu \phi – V(\Abs{\phi}),

where
\label{eqn:qftLecture9:120}
V(\Abs{\phi}) = m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi }^2.

When $$m^2 > 0$$ we have a Wigner mode, but when $$m^2 < 0$$ we have an issue: $$\phi = 0$$ is not a minimum of potential.
When $$m^2 < 0$$ we write
\label{eqn:qftLecture9:140}
\begin{aligned}
V(\phi)
&= – m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi}^2 \\
&= \frac{\lambda}{4} \lr{
\lr{ \phi^\conj \phi}^2 – \frac{4}{\lambda} m^2 } \\
&= \frac{\lambda}{4} \lr{
\phi^\conj \phi – \frac{2}{\lambda} m^2 }^2 – \frac{4 m^4}{\lambda^2},
\end{aligned}

or simply
\label{eqn:qftLecture9:780}
V(\phi)
=
\frac{\lambda}{4} \lr{ \phi^\conj \phi – v^2 }^2 + \text{const}.

The potential (called the Mexican hat potential) is illustrated in fig. 1 for non-zero $$v$$, and in
fig. 2 for $$v = 0$$.
We choose to expand around some point on the minimum ring (it doesn’t matter which one).
When there is no potential, we call the field massless (i.e. if we are in the minimum ring).
We expand as
\label{eqn:qftLecture9:160}
\phi(x) = v \lr{ 1 + \frac{\rho(x)}{v} } e^{i \alpha(x)/v },

so
\label{eqn:qftLecture9:180}
\begin{aligned}
\frac{\lambda}{4}
\lr{\phi^\conj \phi – v^2}^2
&=
\lr{
v^2 \lr{ 1 + \frac{\rho(x)}{v} }^2
– v^2
}^2 \\
&=
\frac{\lambda}{4}
v^4 \lr{ \lr{ 1 + \frac{\rho(x)}{v} }^2 – 1 } \\
&=
\frac{\lambda}{4}
v^4
\lr{
\frac{2 \rho}{v} + \frac{\rho^2}{v^2}
}^2.
\end{aligned}

\label{eqn:qftLecture9:200}
\partial_\mu \phi =
\lr{
v \lr{ 1 + \frac{\rho(x)}{v} } \frac{i}{v} \partial_\mu \alpha
+ \partial_\mu \rho
} e^{i \alpha}

so
\label{eqn:qftLecture9:220}
\begin{aligned}
\LL
&= \Abs{\partial \phi^\conj}^2 – \frac{\lambda}{4} \lr{ \Abs{\phi^\conj}^2 – v^2 }^2 \\
&=
\partial_\mu \rho \partial^\mu \rho + \partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }

\frac{\lambda v^4}{4} \frac{ 4\rho^2}{v^2} + O(\rho^3) \\
&=
\partial_\mu \rho \partial^\mu \rho
– \lambda v^2\rho^2
+
\partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }.
\end{aligned}

We have two fields, $$\rho$$ : a massive scalar field, the “Higgs”, and a massless field $$\alpha$$ (the Goldstone Boson).

$$U(1)$$ symmetry acts on $$\phi(x) \rightarrow e^{i \omega } \phi(x)$$ i.t.o $$\alpha(x) \rightarrow \alpha(x) + v \omega$$.
$$U(1)$$ global symmetry (broken) acts on the Goldstone field $$\alpha(x)$$ by a constant shift. ($$U(1)$$ is still a symmetry of the Lagrangian.)

The current of the $$U(1)$$ symmetry is:
\label{eqn:qftLecture9:240}
j_\mu = \partial_\mu \alpha \lr{ 1 + \text{higher dimensional $$\rho$$ terms} }.

When we quantize
\label{eqn:qftLecture9:260}
\alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp

\label{eqn:qftLecture9:280}
j^\mu(x) = \partial^\mu \alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ i \omega_\Bp – i \Bp } e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ -i \omega_\Bp + i \Bp } e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp.

\label{eqn:qftLecture9:300}
j^\mu(x) \ket{0} \ne 0,

instead it creates a single particle state.

Examples of symmetries

In particle physics, examples of Wigner vs Nambu-Goldstone, ignoring gravity the only exact internal symmetry in the standard module is
$$(B\# – L\#)$$, believed to be a $$U(1)$$ symmetry in Wigner mode.

Here $$B\#$$ is the Baryon number, and $$L\#$$ is the Lepton number. Examples:

• $$B(p) = 1$$, proton.
• $$B(q) = 1/3$$, quark
• $$B(e) = 1$$, electron
• $$B(n) = 1$$, neutron.
• $$L(p) = 1$$, proton.
• $$L(q) = 0$$, quark.
• $$L(e) = 0$$, electron.

The major use of global internal symmetries in the standard model is as “approximate” ones. They become symmetries when one neglects some effect( “terms in $$\LL$$”).
There are other approximate symmetries (use of group theory to find the Balmer series).

Example from HW2:

QCD in limit
\label{eqn:qftLecture9:320}
m_u = m_d = 0.

$$m_u m_d \ll m_p$$ (the products of the up-quark mass and the down-quark mass are much less than a composite one (name?)).
$$SU(2)_L \times SU(2)_R \rightarrow SU(2)_V$$

EWSB (Electro-Weak-Symmetry-Breaking) sector

When the couplings $$g_2, g_1 = 0$$. ($$g_2 \in SU(2), g_1 \in U(1)$$).

Scale invariance

\label{eqn:qftLecture9:340}
\begin{aligned}
x &\rightarrow e^{\lambda} x \\
\phi &\rightarrow e^{-\lambda} \phi \\
A_\mu &\rightarrow e^{-\lambda} A_\mu
\end{aligned}

Any unitary theory which is scale invariant is also \underline{conformal} invariant. Conformal invariance means that angles are preserved.
The point here is that there is more than scale invariance.

We have classical internal global continuous symmetries.
These can be either

1. “unbroken” (Wigner mode)
\label{eqn:qftLecture9:360}
\hatQ\ket{0} = 0.
2. “spontaneously broken”
\label{eqn:qftLecture9:380}
j^\mu(x) \ket{0} \ne 0

(creates Goldstone modes).
3. “anomalous”. Classical symmetries are not a symmetry of QFT.
Examples:

• Scale symmetry (to be studied in QFT II), although this is not truly internal.
• In QCD again when $$\omega_\Bq = 0$$, a $$U(1$$ symmetry (chiral symmetry) becomes exact, and cannot be preserved in QFT.
• In the standard model (E.W sector), the Baryon number and Lepton numbers are not symmetries, but their difference $$B\# – L\#$$ is a symmetry.

Lorentz invariance.

We’d like to study the action of Lorentz symmetries on quantum states. We are going to “go by the book”, finding symmetries, currents, quantize, find generators, and so forth.

Under a Lorentz transformation
\label{eqn:qftLecture9:400}
x^\mu \rightarrow {x’}^\mu = {\Lambda^\mu}_\nu x^\nu,

We are going to consider infinitesimal Lorentz transformations
\label{eqn:qftLecture9:420}
{\Lambda^\mu}_\nu \approx
{\delta^\mu}_\nu + {\omega^\mu}_\nu
,

where $${\omega^\mu}_\nu$$ is small.
A Lorentz transformation $$\Lambda$$ must satisfy $$\Lambda^\T G \Lambda = G$$, or
\label{eqn:qftLecture9:800}
g_{\mu\nu} = {{\Lambda}^\alpha}_\mu g_{\alpha \beta} {{\Lambda}^\beta}_\nu,

into which we insert the infinitesimal transformation representation
\label{eqn:qftLecture9:820}
\begin{aligned}
0
&=
– g_{\mu\nu} +
\lr{ {\delta^\alpha}_\mu + {\omega^\alpha}_\mu }
g_{\alpha \beta}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
\lr{
g_{\mu \beta}
+
\omega_{\beta\mu}
}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
g_{\mu \nu}
+
\omega_{\nu\mu}
+
\omega_{\mu\nu}
+
\omega_{\beta\mu}
{\omega^\beta}_\nu.
\end{aligned}

The quadratic term can be ignored, leaving just
\label{eqn:qftLecture9:840}
0 =
\omega_{\nu\mu}
+
\omega_{\mu\nu},

or
\label{eqn:qftLecture9:860}
\omega_{\nu\mu} = – \omega_{\mu\nu}.

Note that $$\omega$$ is a completely antisymmetric tensor, and like $$F_{\mu\nu}$$ this has only 6 elements.
This means that the
infinitesimal transformation of the coordinates is
\label{eqn:qftLecture9:440}
x^\mu \rightarrow {x’}^\mu \approx x^\mu + \omega^{\mu\nu} x_\nu,

the field transforms as
\label{eqn:qftLecture9:460}
\phi(x) \rightarrow \phi'(x’) = \phi(x)

or
\label{eqn:qftLecture9:760}
\phi'(x^\mu + \omega^{\mu\nu} x_\nu) =
\phi'(x) + \omega^{\mu\nu} x_\nu \partial_\mu\phi(x) = \phi(x),

so
\label{eqn:qftLecture9:480}
\delta \phi = \phi'(x) – \phi(x) =
-\omega^{\mu\nu} x_\nu \partial_\mu \phi.

Since $$\LL$$ is a scalar
\label{eqn:qftLecture9:500}
\begin{aligned}
\delta \LL
&=
-\omega^{\mu\nu} x_\nu \partial_\mu \LL \\
&=

\partial_\mu \lr{
\omega^{\mu\nu} x_\nu \LL
}
+
(\partial_\mu x_\nu) \omega^{\mu\nu} \LL \\
&=
\partial_\mu \lr{

\omega^{\mu\nu} x_\nu \LL
},
\end{aligned}

since $$\partial_\nu x_\mu = g_{\nu\mu}$$ is symmetric, and $$\omega$$ is antisymmetric.
Our current is
\label{eqn:qftLecture9:520}
J^\mu_\omega
=

\omega^{\mu\nu} x_\mu \LL
.

Our Noether current is
\label{eqn:qftLecture9:540}
\begin{aligned}
j^\nu_{\omega^{\mu\rho}}
&= \PD{\phi_{,\nu}}{\LL} \delta \phi – J^\mu_\omega \\
&=
\partial^\nu \phi\lr{ – \omega^{\mu\rho} x_\rho \partial_\mu \phi } + \omega^{\nu \rho} x_\rho \LL \\
&=
\omega^{\mu\rho}
\lr{
\partial^\nu \phi\lr{ – x_\rho \partial_\mu \phi } + {\delta^{\nu}}_\mu x_\rho \LL
} \\
&=
\omega^{\mu\rho} x_\rho
\lr{
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL
}
\end{aligned}

We identify
\label{eqn:qftLecture9:560}

{T^\nu}_\mu =
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL,

so the current is
\label{eqn:qftLecture9:580}
\begin{aligned}
j^\nu_{\omega_{\mu\rho}}
&=
-\omega^{\mu\rho} x_\rho
{T^\nu}_\mu \\
&=
-\omega_{\mu\rho} x^\rho
T^{\nu\mu}
.
\end{aligned}

Define
\label{eqn:qftLecture9:600}
j^{\nu\mu\rho} = \inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} },

which retains the antisymmetry in $$\mu \rho$$ yet still drops the parameter $$\omega^{\mu\rho}$$.
To check that this makes sense, we can contract
$$j^{\nu\mu\rho}$$ with $$\omega_{\rho\mu}$$
\label{eqn:qftLecture9:880}
\begin{aligned}
j^{\nu\mu\rho} \omega_{\rho\mu}
&= -\inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} }
\omega_{\mu\rho} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\mu} T^{\nu\rho}
\omega_{\rho\mu} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\rho} T^{\nu\mu}
\omega_{\mu\rho} \\
&=
– x^{\rho} T^{\nu\mu}
\omega_{\mu\rho},
\end{aligned}

which matches \ref{eqn:qftLecture9:580} as desired.

Example. Rotations $$\mu\rho = ij$$

\label{eqn:qftLecture9:620}
\begin{aligned}
J^{0 i j} \epsilon_{ijk}
&=
\inv{2} \lr{ x^i T^{0j} – x^{j} T^{0i} } \epsilon_{ijk} \\
&=
x^i T^{0j} \epsilon_{ijk}.
\end{aligned}

Observe that this has the structure of $$(\Bx \cross \Bp)_k$$, where $$\Bp$$ is the momentum density of the field.
Let
\label{eqn:qftLecture9:640}
L_k \equiv Q_k = \int d^3 x J^{0ij} \epsilon_{ijk}.

We can now quantize and build a generator
\label{eqn:qftLecture9:660}
\begin{aligned}
\hatU(\Balpha)
&= e^{i \Balpha \cdot \hat{\BL}} \\
&= \exp\lr{i \alpha_k
\int d^3 x x^i \hat{T}^{0j} \epsilon_{ijk}
}
\end{aligned}

From \ref{eqn:qftLecture9:560} we can quantize with $$T^{0j} = \partial^0 \phi \partial^j \phi \rightarrow \hat{\pi} \lr{\spacegrad \phihat}_j$$, or
\label{eqn:qftLecture9:900}
\begin{aligned}
\hatU(\Balpha)
&=
\exp\lr{i \alpha_k
\int d^3 x x^i \hat{\pi} (\spacegrad \phihat)_j \epsilon_{ijk}
} \\
&=
\exp\lr{i \Balpha \cdot
\int d^3 x \hat{\pi} \spacegrad \phihat \cross \Bx
}
\end{aligned}

\label{eqn:qftLecture9:680}
\begin{aligned}
\phihat(\By) \rightarrow \hatU(\alpha) \phihat(\By) \hatU^\dagger(\alpha)
&\approx
\phihat(\By) +
i \Balpha \cdot
\antisymmetric{
\int d^3 x \hat{\pi}(\Bx) \spacegrad \phihat(\Bx) \cross \Bx
}
{
\phihat(\By)
} \\
&=
\phihat(\By) +
i \Balpha \cdot
\int d^3 x
(-i) \delta^3(\Bx – \By)
&=
\phihat(\By) +
\Balpha \cdot \lr{ \spacegrad \phihat(\By ) \cross \By}
\end{aligned}

Explicitly, in coordinates, this is
\label{eqn:qftLecture9:700}
\begin{aligned}
\phihat(\By)
&\rightarrow
\phihat(\By) +
\alpha^i
\lr{
\partial^j \phihat(\By) y^k \epsilon_{jki}
} \\
&=
\phihat(\By) –
\epsilon_{ikj} \alpha^i y^k \partial^j \phihat \\
&=
\phihat( y^j – \epsilon^{ikj} \alpha^i y^k ).
\end{aligned}

This is a rotation. To illustrate, pick $$\Balpha = (0, 0, \alpha)$$, so $$y^j \rightarrow y^j – \epsilon^{ikj} \alpha y^k \delta_{i3} = y^j – \epsilon^{3kj} \alpha y^k$$, or
\label{eqn:qftLecture9:n}
\begin{aligned}
y^1 &\rightarrow y^1 – \epsilon^{3k1} \alpha y^k = y^1 + \alpha y^2 \\
y^2 &\rightarrow y^2 – \epsilon^{3k2} \alpha y^k = y^2 – \alpha y^1 \\
y^3 &\rightarrow y^3 – \epsilon^{3k3} \alpha y^k = y^3,
\end{aligned}

or in matrix form
\label{eqn:qftLecture9:720}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & \alpha & 0 \\
-\alpha & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}.

Totally asymmetric potential

December 16, 2015 phy1520 2 comments , , ,

Q: [1] pr 4.11

(a) Given a time reversal invariant Hamiltonian, show that for any energy eigenket

\label{eqn:totallyAsymmetricPotential:20}
\expectation{\BL} = 0.

(b) If the wave function of such a state is expanded as

\label{eqn:totallyAsymmetricPotential:40}
\sum_{l,m} F_{l m} Y_{l m}(\theta, \phi),

what are the phase restrictions on $$F_{lm}$$?

A: part (a)

For a time reversal invariant Hamiltonian $$H$$ we have

\label{eqn:totallyAsymmetricPotential:60}
H \Theta = \Theta H.

If $$\ket{\psi}$$ is an energy eigenstate with eigenvalue $$E$$, we have

\label{eqn:totallyAsymmetricPotential:80}
\begin{aligned}
H \Theta \ket{\psi}
&= \Theta H \ket{\psi} \\
&= \lambda \Theta \ket{\psi},
\end{aligned}

so $$\Theta \ket{\psi}$$ is also an eigenvalue of $$H$$, so can only differ from $$\ket{\psi}$$ by a phase factor. That is

\label{eqn:totallyAsymmetricPotential:100}
\begin{aligned}
\ket{\psi’}
&=
\Theta \ket{\psi} \\
&= e^{i\delta} \ket{\psi}.
\end{aligned}

Now consider the expectation of $$\BL$$ with respect to a time reversed state

\label{eqn:totallyAsymmetricPotential:120}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\bra{ \psi} \Theta^{-1} \BL \Theta \ket{\psi} \\
&=
\bra{ \psi} (-\BL) \ket{\psi},
\end{aligned}

however, we also have

\label{eqn:totallyAsymmetricPotential:140}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\lr{ \bra{ \psi} e^{-i\delta} } \BL \lr{ e^{i\delta} \ket{\psi} } \\
&=
\bra{\psi} \BL \ket{\psi},
\end{aligned}

so we have $$\bra{\psi} \BL \ket{\psi} = -\bra{\psi} \BL \ket{\psi}$$ which is only possible if $$\expectation{\BL} = \bra{\psi} \BL \ket{\psi} = 0$$.

A: part (b)

Consider the expansion of the wave function of a time reversed energy eigenstate

\label{eqn:totallyAsymmetricPotential:160}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} e^{i\delta} \ket{\psi} \\
&=
e^{i\delta} \braket{\Bx}{\psi},
\end{aligned}

and then consider the same state expanded in the position basis

\label{eqn:totallyAsymmetricPotential:180}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \ket{\Bx’}\bra{\Bx’} } \ket{\psi} \\
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} } \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \Theta \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \ket{\Bx’} \\
&=
\int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \braket{\Bx}{\Bx’} \\
&=
\int d^3 \Bx’ \braket{\psi}{\Bx’} \delta(\Bx- \Bx’) \\
&=
\braket{\psi}{\Bx}.
\end{aligned}

This demonstrates a relationship between the wave function and its complex conjugate

\label{eqn:totallyAsymmetricPotential:200}
\braket{\Bx}{\psi} = e^{-i\delta} \braket{\psi}{\Bx}.

Now expand the wave function in the spherical harmonic basis

\label{eqn:totallyAsymmetricPotential:220}
\begin{aligned}
\braket{\Bx}{\psi}
&=
\int d\Omega \braket{\Bx}{\ncap}\braket{\ncap}{\psi} \\
&=
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) \\
&=
e^{-i\delta}
\lr{
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) }^\conj \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj Y_{lm}^\conj(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj (-1)^m Y_{l,-m}(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{l,-m}(r)}^\conj (-1)^m Y_{l,m}(\theta, \phi),
\end{aligned}

so the $$F_{lm}$$ functions are constrained by

\label{eqn:totallyAsymmetricPotential:240}
F_{lm}(r) = e^{-i\delta} \lr{ F_{l,-m}(r)}^\conj (-1)^m.

References

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

Angular momentum expectation values

December 14, 2015 phy1520 No comments ,

Q: [1] pr 3.18

Compute the expectation values for the first and second powers of the angular momentum operators with respect to states $$\ket{lm}$$.

A:

We can write the expectation values for the $$L_z$$ powers immediately

\label{eqn:angularMomentumExpectation:20}
\expectation{L_z}
= m \Hbar,

and

\label{eqn:angularMomentumExpectation:40}
\expectation{L_z^2} = (m \Hbar)^2.

For the x and y components first express the operators in terms of the ladder operators.

\label{eqn:angularMomentumExpectation:60}
\begin{aligned}
L_{+} &= L_x + i L_y \\
L_{-} &= L_x – i L_y.
\end{aligned}

Rearranging gives

\label{eqn:angularMomentumExpectation:80}
\begin{aligned}
L_x &= \inv{2} \lr{ L_{+} + L_{-} } \\
L_y &= \inv{2i} \lr{ L_{+} – L_{-} }.
\end{aligned}

The first order expectations $$\expectation{L_x}, \expectation{L_y}$$ are both zero since $$\expectation{L_{+}} = \expectation{L_{-}}$$. For the second order expectation values we have

\label{eqn:angularMomentumExpectation:100}
\begin{aligned}
L_x^2
&= \inv{4} \lr{ L_{+} + L_{-} } \lr{ L_{+} + L_{-} } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + L_{+} L_{-} + L_{-} L_{+} } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (L_x^2 + L_y^2) } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (\BL^2 – L_z^2) },
\end{aligned}

and
\label{eqn:angularMomentumExpectation:120}
\begin{aligned}
L_y^2
&= -\inv{4} \lr{ L_{+} – L_{-} } \lr{ L_{+} – L_{-} } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – L_{+} L_{-} – L_{-} L_{+} } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (L_x^2 + L_y^2) } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (\BL^2 – L_z^2) }.
\end{aligned}

Any expectation value $$\bra{lm} L_{+} L_{+} \ket{lm}$$ or $$\bra{lm} L_{-} L_{-} \ket{lm}$$ will be zero, leaving

\label{eqn:angularMomentumExpectation:140}
\begin{aligned}
\expectation{L_x^2}
&=
\expectation{L_y^2} \\
&=
\inv{4} \expectation{2 (\BL^2 – L_z^2) } \\
&=
\inv{2} \lr{ \Hbar^2 l(l+1) – (\Hbar m)^2 }.
\end{aligned}

Observe that we have
\label{eqn:angularMomentumExpectation:160}
\expectation{L_x^2}
+
\expectation{L_y^2}
+
\expectation{L_z^2}
=
\Hbar^2 l(l+1)
=
\expectation{\BL^2},

which is the quantum mechanical analogue of the classical scalar equation $$\BL^2 = L_x^2 + L_y^2 + L_z^2$$.

References

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

L_z and L^2 eigenvalues and probabilities for a given wave function

Q: [1] 3.17

Given a wave function

\label{eqn:LsquaredLzProblem:20}
\psi(r,\theta, \phi) = f(r) \lr{ x + y + 3 z },

• (a) Determine if this wave function is an eigenfunction of $$\BL^2$$, and the value of $$l$$ if it is an eigenfunction.

• (b) Determine the probabilities for the particle to be found in any given $$\ket{l, m}$$ state,
• (c) If it is known that $$\psi$$ is an energy eigenfunction with energy $$E$$ indicate how we can find $$V(r)$$.

A: (a)

Using
\label{eqn:LsquaredLzProblem:40}
\BL^2
=
-\Hbar^2 \lr{ \inv{\sin^2\theta} \partial_{\phi\phi} + \inv{\sin\theta} \partial_\theta \lr{ \sin\theta \partial_\theta} },

and

\label{eqn:LsquaredLzProblem:60}
\begin{aligned}
x &= r \sin\theta \cos\phi \\
y &= r \sin\theta \sin\phi \\
z &= r \cos\theta
\end{aligned}

it’s a quick computation to show that

\label{eqn:LsquaredLzProblem:80}
\BL^2 \psi = 2 \Hbar^2 \psi = 1(1 + 1) \Hbar^2 \psi,

so this function is an eigenket of $$\BL^2$$ with an eigenvalue of $$2 \Hbar^2$$, which corresponds to $$l = 1$$, a p-orbital state.

(b)

Recall that the angular representation of $$L_z$$ is

\label{eqn:LsquaredLzProblem:100}
L_z = -i \Hbar \PD{\phi},

so we have

\label{eqn:LsquaredLzProblem:120}
\begin{aligned}
L_z x &= i \Hbar y \\
L_z y &= – i \Hbar x \\
L_z z &= 0,
\end{aligned}

The $$L_z$$ action on $$\psi$$ is

\label{eqn:LsquaredLzProblem:140}
L_z \psi = -i \Hbar r f(r) \lr{ – y + x }.

This wave function is not an eigenket of $$L_z$$. Expressed in terms of the $$L_z$$ basis states $$e^{i m \phi}$$, this wave function is

\label{eqn:LsquaredLzProblem:160}
\begin{aligned}
\psi
&= r f(r) \lr{ \sin\theta \lr{ \cos\phi + \sin\phi} + \cos\theta } \\
&= r f(r) \lr{ \frac{\sin\theta}{2} \lr{ e^{i \phi} \lr{ 1 + \inv{i}} + e^{-i\phi} \lr{ 1 – \inv{i} } } + \cos\theta } \\
&= r f(r) \lr{
\frac{(1-i)\sin\theta}{2} e^{1 i \phi}
+
\frac{(1+i)\sin\theta}{2} e^{- 1 i \phi}
+ \cos\theta e^{0 i \phi}
}
\end{aligned}

Assuming that $$\psi$$ is normalized, the probabilities for measuring $$m = 1,-1,0$$ respectively are

\label{eqn:LsquaredLzProblem:180}
\begin{aligned}
P_{\pm 1}
&= 2 \pi \rho \Abs{\frac{1\mp i}{2}}^2 \int_0^\pi \sin\theta d\theta \sin^2 \theta \\
&= -2 \pi \rho \int_1^{-1} du (1-u^2) \\
&= 2 \pi \rho \evalrange{ \lr{ u – \frac{u^3}{3} } }{-1}{1} \\
&= 2 \pi \rho \lr{ 2 – \frac{2}{3}} \\
&= \frac{ 8 \pi \rho}{3},
\end{aligned}

and

\label{eqn:LsquaredLzProblem:200}
P_{0} = 2 \pi \rho \int_0^\pi \sin\theta \cos\theta = 0,

where

\label{eqn:LsquaredLzProblem:220}
\rho = \int_0^\infty r^4 \Abs{f(r)}^2 dr.

Because the probabilities must sum to 1, this means the $$m = \pm 1$$ states are equiprobable with $$P_{\pm} = 1/2$$, fixing $$\rho = 3/16\pi$$, even without knowing $$f(r)$$.

(c)

The operator $$r^2 \Bp^2$$ can be decomposed into a $$\BL^2$$ component and some other portions, from which we can write

\label{eqn:LsquaredLzProblem:240}
\begin{aligned}
H \psi
&= \lr{ \frac{\Bp^2}{2m} + V(r) } \psi \\
&=
\lr{
– \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \inv{\Hbar^2 r^2} \BL^2 } + V(r) } \psi.
\end{aligned}

(See: [1] eq. 6.21)

In this case where $$\BL^2 \psi = 2 \Hbar^2 \psi$$ we can rearrange for $$V(r)$$

\label{eqn:LsquaredLzProblem:260}
\begin{aligned}
V(r)
&= E + \inv{\psi} \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \frac{2}{r^2} } \psi \\
&= E + \inv{f(r)} \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \frac{2}{r^2} } f(r).
\end{aligned}

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 15: angular momentum rotation representation, and angular momentum addition. 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 chap 3. content from [1].

Angular momentum (wrap up.)

We found

\label{eqn:qmLecture15:20}
\begin{aligned}
\hat{\BL^2} \ket{j, m} &= j(j+1) \Hbar^2 \ket{j,m} \\
\hat{L}_z \ket{j, m} &= \Hbar m \ket{j,m} \\
\hat{L}_{\pm} \ket{j, m } &= \Hbar \sqrt{(j \mp m)(j \pm m + 1)} \ket{j, m \pm 1 }
\end{aligned}

or Schwinger

\label{eqn:qmLecture15:40}
\begin{aligned}
\hat{L}_z &= \inv{2} \lr{ \hat{n}_1 – \hat{n}_2 } \Hbar \\
\hat{L}_{+} &= a_1^\dagger a_2 \Hbar \\
\hat{L}_{-} &= a_1 a_2^\dagger \Hbar \\
j &= \inv{2} \lr{ \hat{n}_1 + \hat{n}_2 },
\end{aligned}

where each of the $$a_1, a_2$$ operators obey

\label{eqn:qmLecture15:60}
\begin{aligned}
\antisymmetric{a_1}{a_1^\dagger} &= 1 \\
\antisymmetric{a_2}{a_2^\dagger} &= 1
\end{aligned}

and any pair of different index $$a$$ operators commute, as in

\label{eqn:qmLecture15:80}
\antisymmetric{a_1}{a_2^\dagger} = 0.

Representations

It’s possible to compute matrix representations of the rotation operators

\label{eqn:qmLecture15:100}
\hat{R}_\ncap(\phi) = e^{i \hat{\BL} \cdot \ncap \phi/\Hbar}.

With respect to a ket it’s possible to find

\label{eqn:qmLecture15:120}
e^{i \hat{\BL} \cdot \ncap \phi/\Hbar} \ket{j, m}
=
\sum_{m’} d^j_{m m’}(\ncap, \phi) \ket{ j, m’ }.

This has a block diagonal form that’s sketched in fig. 1.

fig. 1. Block diagonal form for angular momentum matrix representation.

We can view $$d^j_{m m’}(\ncap, \phi)$$ as a matrix, representing the rotation. The problem of determining these matrices can be reduced to that of determining the matrix for $$\hat{\BL}$$, because once we have that we can exponentiate that.

Example: spin 1/2

From the eigenvalue relationships, with basis states

\label{eqn:qmLecture15:160}
\begin{aligned}
\ket{\uparrow} &=
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
\ket{\downarrow} &=
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix}
\end{aligned}

we find

\label{eqn:qmLecture15:180}
\begin{aligned}
\hat{L}_z &= \frac{\Hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\
\hat{L}_{+} &= \frac{\Hbar}{2}
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} \\
\hat{L}_{-} &= \frac{\Hbar}{2}
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}.
\end{aligned}

Rearranging we find the Pauli matrices

\label{eqn:qmLecture15:200}
\hat{L}_k = \inv{2} \Hbar \sigma_i.

Noting that $$\lr{ \Bsigma \cdot \ncap }^2 = 1$$, and $$\lr{\Bsigma \cdot \ncap }^3 = \Bsigma \cdot \ncap$$, the rotation matrix is

\label{eqn:qmLecture15:220}
e^{ i \Bsigma \cdot \ncap \phi/2 } \ket{\inv{2}, m} = \lr{ \cos( \phi/2 ) + i \Bsigma \cdot \ncap \sin(\phi/2) } \ket{\inv{2}, m}.

The steps are

1. Enumerate the states.
\label{eqn:qmLecture15:140}
j_1 = \inv{2} \leftrightarrow\, \mbox{2 states (dimension of irrep = 2)}

2. Construct the $$\hat{\BL}$$ matrices.
3. Construct $$d_{m m’}^j(\ncap, \phi)$$.

Angular momentum operator in space representation

For $$L = 1$$ it turns out that the rotation matrices turn out to be the 3D rotation matrices. In the space representation

\label{eqn:qmLecture15:240}
\BL = \Br \cross \Bp,

the coordinates of the operator are

\label{eqn:qmLecture15:260}
\hat{L}_k = i \epsilon_{k m n} r_m \lr{ -i \Hbar \PD{r_n}{} }

We see that scaling $$\Br \rightarrow \alpha \Br$$ does not change this operator, allowing for an angular representation $$\hat{L}_k(\theta, \phi)$$ that have the form

\label{eqn:qmLecture15:280}
\begin{aligned}
\hat{L}_z &= -i \Hbar \PD{\phi}{} \\
\hat{L}_{\pm} &= \Hbar \lr{ \pm \PD{\theta}{} + i \cot \theta \PD{\phi}{} }.
\end{aligned}

Here $$\theta$$ and $$\phi$$ are the polar and azimuthal angles respectively as illustrated in fig. 2.

fig. 2. Spherical coordinate convention.

The equivalent wave function representation of the problem is

\label{eqn:qmLecture15:300}
\begin{aligned}
\hat{\BL} Y_{lm}(\theta, \phi) &= \Hbar^2 l (l + 1) Y_{lm}(\theta, \phi) \\
\hat{L}_z Y_{lm}(\theta, \phi) &= \Hbar m Y_{lm}(\theta, \phi) \\
\end{aligned}

One can find these functions

\label{eqn:qmLecture15:320}
Y_{lm}(\theta, \phi) = P_{l, m}(\cos \theta) e^{i m \phi},

where $$P_{l, m}(\cos \theta)$$ are called the associated Legendre polynomials. This can be applied whenever we have

\label{eqn:qmLecture15:340}
\antisymmetric{H}{\hat{L}_k} = 0.

where all the eigenfunctions will have the form

\label{eqn:qmLecture15:360}
\Psi(r, \theta, \phi) = R(r) Y_{lm}(\theta, \phi).

Since $$\hat{\BL}$$ is a vector we expect to be able to add angular momentum in some way similar to the addition of classical vectors as illustrated in fig. 3.

When we have a potential that depends only on the difference in position $$V(\Br_1 – \Br_2)$$ then we know from classical problems it is effective to work in center of mass coordinates

\label{eqn:qmLecture15:380}
\begin{aligned}
\hat{\BR}_{\textrm{cm}} &= \frac{\hat{\Br}_1 + \hat{\Br}_2}{2} \\
\hat{\BP}_{\textrm{cm}} &= \hat{\Bp}_1 + \hat{\Bp}_2
\end{aligned}

where

\label{eqn:qmLecture15:400}
\antisymmetric{\hat{R}_i}{\hat{P}_j} = i \Hbar \delta_{ij}.

Given

\label{eqn:qmLecture15:420}
\hat{\BL}_1 + \hat{\BL}_2 = \hat{\BL}_{\textrm{tot}},

do we have
\label{eqn:qmLecture15:440}
\antisymmetric{
\hat{L}_{\textrm{tot}, i}
}{
\hat{L}_{\textrm{tot}, j}
}
= i \Hbar \epsilon_{i j k} \hat{L}_{\textrm{tot}, k} ?

That is

\label{eqn:qmLecture15:460}
\antisymmetric{\hat{L}_{1,i} + \hat{L}_{1,j}}{\hat{L}_{2,i} + \hat{L}_{2,j}} = i \Hbar \epsilon_{i j k} \lr{ \hat{L}_{1,k} + \hat{L}_{1,k} }

FIXME: Right at the end of the lecture, there was a mention of something about whether or not $$\hat{\BL}_1^2$$ and $$\hat{L}_{1,z}$$ were sharply defined, but I missed it. Ask about this if not covered in the next lecture.

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.

Third update of aggregate notes for phy1520, Graduate Quantum Mechanics.

I’ve posted a third update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 13, my solutions for the third problem set, and some additional worked practice problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

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.

Commutators of angular momentum and a central force Hamiltonian

September 30, 2015 phy1520 No comments , , , ,

In problem 1.17 of [1] we are to show that non-commuting operators that both commute with the Hamiltonian, have, in general, degenerate energy eigenvalues. It suggests considering $$L_x, L_z$$ and a central force Hamiltonian $$H = \Bp^2/2m + V(r)$$ as examples.

Let’s just demonstrate these commutators act as expected in these cases.

With $$\BL = \Bx \cross \Bp$$, we have

\label{eqn:angularMomentumAndCentralForceCommutators:20}
\begin{aligned}
L_x &= y p_z – z p_y \\
L_y &= z p_x – x p_z \\
L_z &= x p_y – y p_x.
\end{aligned}

The $$L_x, L_z$$ commutator is

\label{eqn:angularMomentumAndCentralForceCommutators:40}
\begin{aligned}
\antisymmetric{L_x}{L_z}
&=
\antisymmetric{y p_z – z p_y }{x p_y – y p_x} \\
&=
\antisymmetric{y p_z}{x p_y}
-\antisymmetric{y p_z}{y p_x}
-\antisymmetric{z p_y }{x p_y}
+\antisymmetric{z p_y }{y p_x} \\
&=
x p_z \antisymmetric{y}{p_y}
+ z p_x \antisymmetric{p_y }{y} \\
&=
i \Hbar \lr{ x p_z – z p_x } \\
&=
– i \Hbar L_y
\end{aligned}

cyclicly permuting the indexes shows that no pairs of different $$\BL$$ components commute. For $$L_y, L_x$$ that is

\label{eqn:angularMomentumAndCentralForceCommutators:60}
\begin{aligned}
\antisymmetric{L_y}{L_x}
&=
\antisymmetric{z p_x – x p_z }{y p_z – z p_y} \\
&=
\antisymmetric{z p_x}{y p_z}
-\antisymmetric{z p_x}{z p_y}
-\antisymmetric{x p_z }{y p_z}
+\antisymmetric{x p_z }{z p_y} \\
&=
y p_x \antisymmetric{z}{p_z}
+ x p_y \antisymmetric{p_z }{z} \\
&=
i \Hbar \lr{ y p_x – x p_y } \\
&=
– i \Hbar L_z,
\end{aligned}

and for $$L_z, L_y$$

\label{eqn:angularMomentumAndCentralForceCommutators:80}
\begin{aligned}
\antisymmetric{L_z}{L_y}
&=
\antisymmetric{x p_y – y p_x }{z p_x – x p_z} \\
&=
\antisymmetric{x p_y}{z p_x}
-\antisymmetric{x p_y}{x p_z}
-\antisymmetric{y p_x }{z p_x}
+\antisymmetric{y p_x }{x p_z} \\
&=
z p_y \antisymmetric{x}{p_x}
+ y p_z \antisymmetric{p_x }{x} \\
&=
i \Hbar \lr{ z p_y – y p_z } \\
&=
– i \Hbar L_x.
\end{aligned}

If these angular momentum components are also shown to commute with themselves (which they do), the commutator relations above can be summarized as

\label{eqn:angularMomentumAndCentralForceCommutators:100}
\antisymmetric{L_a}{L_b} = i \Hbar \epsilon_{a b c} L_c.

In the example to consider, we’ll have to consider the commutators with $$\Bp^2$$ and $$V(r)$$. Picking any one component of $$\BL$$ is sufficent due to the symmetries of the problem. For example

\label{eqn:angularMomentumAndCentralForceCommutators:120}
\begin{aligned}
\antisymmetric{L_x}{\Bp^2}
&=
\antisymmetric{y p_z – z p_y}{p_x^2 + p_y^2 + p_z^2} \\
&=
\antisymmetric{y p_z}{{p_x^2} + p_y^2 + {p_z^2}}
-\antisymmetric{z p_y}{{p_x^2} + {p_y^2} + p_z^2} \\
&=
p_z \antisymmetric{y}{p_y^2}
-p_y \antisymmetric{z}{p_z^2} \\
&=
p_z 2 i \Hbar p_y
2 i \Hbar p_y
-p_y 2 i \Hbar p_z \\
&=
0.
\end{aligned}

How about the commutator of $$\BL$$ with the potential? It is sufficient to consider one component again, for example

\label{eqn:angularMomentumAndCentralForceCommutators:140}
\begin{aligned}
\antisymmetric{L_x}{V}
&=
\antisymmetric{y p_z – z p_y}{V} \\
&=
y \antisymmetric{p_z}{V} – z \antisymmetric{p_y}{V} \\
&=
-i \Hbar y \PD{z}{V(r)} + i \Hbar z \PD{y}{V(r)} \\
&=
-i \Hbar y \PD{r}{V}\PD{z}{r} + i \Hbar z \PD{r}{V}\PD{y}{r} \\
&=
-i \Hbar y \PD{r}{V} \frac{z}{r} + i \Hbar z \PD{r}{V}\frac{y}{r} \\
&=
0.
\end{aligned}

We’ve shown that all the components of $$\BL$$ commute with a central force Hamiltonian, and each different component of $$\BL$$ do not commute.

The next step will be figuring out how to use this to show that there are energy degeneracies.

References

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

Update to old phy356 (Quantum Mechanics I) notes.

It’s been a long time since I took QM I. My notes from that class were pretty rough, but I’ve cleaned them up a bit.

The main value to these notes is that I worked a number of introductory Quantum Mechanics problems.

These were my personal lecture notes for the Fall 2010, University of Toronto Quantum mechanics I course (PHY356H1F), taught by Prof. Vatche Deyirmenjian.

The official description of this course was:

The general structure of wave mechanics; eigenfunctions and eigenvalues; operators; orbital angular momentum; spherical harmonics; central potential; separation of variables, hydrogen atom; Dirac notation; operator methods; harmonic oscillator and spin.

This document contains a few things

• My lecture notes.
Typos, if any, are probably mine(Peeter), and no claim nor attempt of spelling or grammar correctness will be made. The first four lectures had chosen not to take notes for since they followed the text very closely.
• Notes from reading of the text. This includes observations, notes on what seem like errors, and some solved problems. None of these problems have been graded. Note that my informal errata sheet for the text has been separated out from this document.
• Some assigned problems. I have corrected some the errors after receiving grading feedback, and where I have not done so I at least recorded some of the grading comments as a reference.
• Some worked problems associated with exam preparation.