## Time reversal behavior of solutions to crystal spin Hamiltonian

December 15, 2015 phy1520 , , , ,

### Q: [1] pr 4.12

Solve the spin 1 Hamiltonian
\label{eqn:crystalSpinHamiltonianTimeReversal:20}
H = A S_z^2 + B(S_x^2 – S_y^2).

Is this Hamiltonian invariant under time reversal?

How do the eigenkets change under time reversal?

In spinMatrices.nb the matrix representation of the Hamiltonian is found to be
\label{eqn:crystalSpinHamiltonianTimeReversal:40}
H =
\Hbar^2
\begin{bmatrix}
A & 0 & B \\
0 & 0 & 0 \\
B & 0 & A
\end{bmatrix}.

The eigenvalues are
\label{eqn:crystalSpinHamiltonianTimeReversal:60}
\setlr{ 0, A – B, A + B},

and the respective eigenvalues (unnormalized) are

\label{eqn:crystalSpinHamiltonianTimeReversal:80}
\setlr{
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
1 \\
0 \\
1 \\
\end{bmatrix}
}.

Under time reversal, the Hamiltonian is

\label{eqn:crystalSpinHamiltonianTimeReversal:100}
H \rightarrow A (-S_z)^2 + B ( (-S_x)^2 – (-S_y)^2 ) = H,

so we expect the eigenkets for this Hamiltonian to vary by at most a phase factor. To check this, first recall that the time reversal action on a spin one state is

\label{eqn:crystalSpinHamiltonianTimeReversal:120}
\Theta \ket{1, m} = (-1)^m \ket{1, -m},

or

\label{eqn:crystalSpinHamiltonianTimeReversal:140}
\begin{aligned}
\Theta \ket{1,1} &= -\ket{1,-1} \\
\Theta \ket{1,0} &= \ket{1,0} \\
\Theta \ket{1,-1} &= -\ket{1,1}.
\end{aligned}

Let’s write the eigenkets respectively as

\label{eqn:crystalSpinHamiltonianTimeReversal:160}
\begin{aligned}
\ket{0} &= \ket{1,0} \\
\ket{A-B} &= -\ket{1,-1} + \ket{1,1} \\
\ket{A+B} &= \ket{1,-1} + \ket{1,1}.
\end{aligned}

Under the reversal operation, we should have

\label{eqn:crystalSpinHamiltonianTimeReversal:180}
\begin{aligned}
\Theta \ket{0} &\rightarrow \ket{1,0} \\
\Theta \ket{A-B} &= +\ket{1,-1} – \ket{1,1} \\
\Theta \ket{A+B} &= -\ket{1,-1} – \ket{1,1}.
\end{aligned}

Up to a sign, the time reversed states match the unreversed states, which makes sense given the Hamiltonian invariance.

# References

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

## Spin three halves spin interaction

December 15, 2015 phy1520 , , , ,

### Q: [1] pr 3.33

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

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

Show that the interaction energy can be written as

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

Find the energy eigenvalues for such a Hamiltonian.

### A:

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

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

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

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

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

For a static electric field we have

\label{eqn:spinThreeHalvesNucleus:160}

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

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

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

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

### A: energy eigenvalues

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

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

The energy eigenvalues are found to be

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

with two fold degeneracies for each eigenvalue.

# References

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

## Angular momentum expectation values

December 14, 2015 phy1520 ,

### 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.

## Expectation of spherically symmetric 3D potential derivative

### Q: [1] pr 5.16

For a particle in a spherically symmetric potential $$V(r)$$ show that

\label{eqn:symmetricPotentialDerivativeExpectation:20}
\Abs{\psi(0)}^2 = \frac{m}{2 \pi \Hbar^2} \expectation{ \frac{dV}{dr} },

for all s-states, ground or excited.

Then show this is the case for the 3D SHO and hydrogen wave functions.

### A:

The text works a problem that looks similar to this by considering the commutator of an operator $$A$$, later set to $$A = p_r = -i \Hbar \PDi{r}{}$$ the radial momentum operator. First it is noted that

\label{eqn:symmetricPotentialDerivativeExpectation:40}
0 = \bra{nlm} \antisymmetric{H}{A} \ket{nlm},

since $$H$$ operating to either the right or the left is the energy eigenvalue $$E_n$$. Next it appears the author uses an angular momentum factoring of the squared momentum operator. Looking earlier in the text that factoring is found to be

\label{eqn:symmetricPotentialDerivativeExpectation:60}
\frac{\Bp^2}{2m}
= \inv{2 m r^2} \BL^2 – \frac{\Hbar^2}{2m} \lr{ \PDSq{r}{} + \frac{2}{r} \PD{r}{} }.

With
\label{eqn:symmetricPotentialDerivativeExpectation:80}
R = – \frac{\Hbar^2}{2m} \lr{ \PDSq{r}{} + \frac{2}{r} \PD{r}{} }.

we have

\label{eqn:symmetricPotentialDerivativeExpectation:100}
\begin{aligned}
0
&= \bra{nlm} \antisymmetric{H}{p_r} \ket{nlm} \\
&= \bra{nlm} \antisymmetric{\frac{\Bp^2}{2m} + V(r)}{p_r} \ket{nlm} \\
&= \bra{nlm} \antisymmetric{\inv{2 m r^2} \BL^2 + R + V(r)}{p_r} \ket{nlm} \\
&= \bra{nlm} \antisymmetric{\frac{-\Hbar^2 l (l+1)}{2 m r^2} + R + V(r)}{p_r} \ket{nlm}.
\end{aligned}

Let’s consider the commutator of each term separately. First

\label{eqn:symmetricPotentialDerivativeExpectation:120}
\begin{aligned}
\antisymmetric{V}{p_r} \psi
&=
V p_r \psi

p_r V \psi \\
&=
V p_r \psi

(p_r V) \psi

V p_r \psi \\
&=

(p_r V) \psi \\
&=
i \Hbar \PD{r}{V} \psi.
\end{aligned}

Setting $$V(r) = 1/r^2$$, we also have

\label{eqn:symmetricPotentialDerivativeExpectation:160}
\antisymmetric{\inv{r^2}}{p_r} \psi
=
-\frac{2 i \Hbar}{r^3} \psi.

Finally
\label{eqn:symmetricPotentialDerivativeExpectation:180}
\begin{aligned}
\antisymmetric{\PDSq{r}{} + \frac{2}{r} \PD{r}{} }{ \PD{r}{}}
&=
\lr{ \partial_{rr} + \frac{2}{r} \partial_r } \partial_r

\partial_r \lr{ \partial_{rr} + \frac{2}{r} \partial_r } \\
&=
\partial_{rrr} + \frac{2}{r} \partial_{rr}

\lr{
\partial_{rrr} -\frac{2}{r^2} \partial_r + \frac{2}{r} \partial_{rr}
} \\
&=
-\frac{2}{r^2} \partial_r,
\end{aligned}

so
\label{eqn:symmetricPotentialDerivativeExpectation:200}
\antisymmetric{R}{p_r}
=-\frac{2}{r^2} \frac{-\Hbar^2}{2m} p_r
=\frac{\Hbar^2}{m r^2} p_r.

Putting all the pieces back together, we’ve got
\label{eqn:symmetricPotentialDerivativeExpectation:220}
\begin{aligned}
0
&= \bra{nlm} \antisymmetric{\frac{-\Hbar^2 l (l+1)}{2 m r^2} + R + V(r)}{p_r} \ket{nlm} \\
&=
i \Hbar
\bra{nlm} \lr{
\frac{\Hbar^2 l (l+1)}{m r^3} – \frac{i\Hbar}{m r^2} p_r +
\PD{r}{V}
}
\ket{nlm}.
\end{aligned}

Since s-states are those for which $$l = 0$$, this means

\label{eqn:symmetricPotentialDerivativeExpectation:240}
\begin{aligned}
\expectation{\PD{r}{V}}
&= \frac{i\Hbar}{m } \expectation{ \inv{r^2} p_r } \\
&= \frac{\Hbar^2}{m } \expectation{ \inv{r^2} \PD{r}{} } \\
&= \frac{\Hbar^2}{m } \int_0^\infty dr \int_0^\pi d\theta \int_0^{2 \pi} d\phi r^2 \sin\theta \psi^\conj(r,\theta, \phi) \inv{r^2} \PD{r}{\psi(r,\theta,\phi)}.
\end{aligned}

Since s-states are spherically symmetric, this is
\label{eqn:symmetricPotentialDerivativeExpectation:260}
\expectation{\PD{r}{V}}
= \frac{4 \pi \Hbar^2}{m } \int_0^\infty dr \psi^\conj \PD{r}{\psi}.

That integral is

\label{eqn:symmetricPotentialDerivativeExpectation:280}
\int_0^\infty dr \psi^\conj \PD{r}{\psi}
=
\evalrange{\Abs{\psi}^2}{0}{\infty} – \int_0^\infty dr \PD{r}{\psi^\conj} \psi.

With the hydrogen atom, our radial wave functions are real valued. It’s reasonable to assume that we can do the same for other real-valued spherical potentials. If that is the case, we have

\label{eqn:symmetricPotentialDerivativeExpectation:300}
2 \int_0^\infty dr \psi^\conj \PD{r}{\psi}
=
\Abs{\psi(0)}^2,

and

\label{eqn:symmetricPotentialDerivativeExpectation:320}
\boxed{
\expectation{\PD{r}{V}}
= \frac{2 \pi \Hbar^2}{m } \Abs{\psi(0)}^2,
}

which completes this part of the problem.

### A: show this is the case for the 3D SHO and hydrogen wave functions

For a hydrogen like atom, in atomic units, we have

\label{eqn:symmetricPotentialDerivativeExpectation:360}
\begin{aligned}
\expectation{
\PD{r}{V}
}
&=
\expectation{
\PD{r}{} \lr{ -\frac{Z e^2}{r} }
} \\
&=
Z e^2
\expectation
{
\inv{r^2}
} \\
&=
Z e^2 \frac{Z^2}{n^3 a_0^2 \lr{ l + 1/2 }} \\
&=
\frac{\Hbar^2}{m a_0} \frac{2 Z^3}{n^3 a_0^2} \\
&=
\frac{2 \Hbar^2 Z^3}{m n^3 a_0^3}.
\end{aligned}

On the other hand for $$n = 1$$, we have

\label{eqn:symmetricPotentialDerivativeExpectation:380}
\begin{aligned}
\frac{2 \pi \Hbar^2}{m} \Abs{R_{10}(0)}^2 \Abs{Y_{00}}^2
&=
\frac{2 \pi \Hbar^2}{m} \frac{Z^3}{a_0^3} 4 \inv{4 \pi} \\
&=
\frac{2 \Hbar^2 Z^3}{m a_0^3},
\end{aligned}

and for $$n = 2$$, we have

\label{eqn:symmetricPotentialDerivativeExpectation:400}
\begin{aligned}
\frac{2 \pi \Hbar^2}{m} \Abs{R_{20}(0)}^2 \Abs{Y_{00}}^2
&=
\frac{2 \pi \Hbar^2}{m} \frac{Z^3}{8 a_0^3} 4 \inv{4 \pi} \\
&=
\frac{\Hbar^2 Z^3}{4 m a_0^3}.
\end{aligned}

These both match the potential derivative expectation when evaluated for the s-orbital ($$l = 0$$).

For the 3D SHO I verified the ground state case in the Mathematica notebook sakuraiProblem5.16bSHO.nb

There it was found that

\label{eqn:symmetricPotentialDerivativeExpectation:420}
\expectation{\PD{r}{V}}
= \frac{2 \pi \Hbar^2}{m } \Abs{\psi(0)}^2
= 2 \sqrt{\frac{m \omega ^3 \Hbar}{ \pi }}

# References

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

### Q: [1] pr. 5.18

Work out the quadratic Zeeman effect for the ground state hydrogen atom due to the usually neglected $$e^2 \BA^2/2 m_e c^2$$ term in the Hamiltonian.

### A:

The first order energy shift is

For a z-oriented magnetic field we can use

\BA = \frac{B}{2} \setlr{ -y, x, 0 },

so the perturbation potential is

\begin{aligned}
V
&= \frac{e^2 \BA^2}{2 m_e c^2} \\
&= \frac{e^2 \BB^2 (x^2 + y^2)}{8 m_e c^2} \\
&= \frac{ e^2 \BB^2 r^2 \sin^2\theta }{8 m_e c^2}
\end{aligned}

The ground state wave function is

\begin{aligned}
\psi_0
&= \braket{\Bx}{0} \\
&= \inv{\sqrt{\pi a_0^3}} e^{-r/a_0},
\end{aligned}

so the energy shift is

\begin{aligned}
\Delta
&= \bra{0} V \ket{0} \\
&= \inv{ \pi a_0^3 } 2 \pi \frac{ e^2 \BB^2 }{8 m_e c^2} \int_0^\infty r^2 \sin\theta e^{-2r/a_0} r^2 \sin^2\theta dr d\theta \\
&=
\frac{ e^2 \BB^2 }{4 a_0^3 m_e c^2}
\int_0^\infty r^4 e^{-2r/a_0} dr \int_0^\pi \sin^3\theta d\theta \\
&= –
\frac{ e^2 \BB^2 }{4 a_0^3 m_e c^2}
\frac{4!}{(2/a_0)^{4+1} } \evalrange{\lr{u – \frac{u^3}{3}}}{1}{-1} \\
&=
\frac{ e^2 a_0^2 \BB^2 }{4 m_e c^2}.
\end{aligned}

If this energy shift is written in terms of a diamagnetic susceptibility $$\chi$$ defined by

\Delta = -\inv{2} \chi \BB^2,

the diamagnetic susceptibility is

\chi = -\frac{ e^2 a_0^2 }{2 m_e c^2}.

# References

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