## 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.

## PHY1520H Graduate Quantum Mechanics. Lecture 22: Van der Wall potential and Stark effect. Taught by Prof. Arun Paramekanti

December 10, 2015 phy1520 , , ,

### 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] ch. 5 content.

### Another approach (for last time?)

Imagine we perturb a potential, say a harmonic oscillator with an electric field

\label{eqn:qmLecture22:20}
V_0(x) = \inv{2} k x^2

\label{eqn:qmLecture22:40}
V(x) = \mathcal{E} e x

After minimizing the energy, using $$\PDi{x}{V} = 0$$, we get

\label{eqn:qmLecture22:60}
\inv{2} k x^2 + \mathcal{E} e x \rightarrow k x^\conj = – e \mathcal{E}

\label{eqn:qmLecture22:80}
p^\conj = -e x^\conj = – \frac{e^2 \mathcal{E}}{k}

For such a system the polarizability is

\label{eqn:qmLecture22:100}
\alpha = \frac{e^2 }{k}

\label{eqn:qmLecture22:120}
\begin{aligned}
\inv{2} k \lr{ -\frac{ e \mathcal{E}}{k} }^2 + \mathcal{E} e \lr{ – \frac{e \mathcal{E}}{k} }
&= – \inv{2} \lr{ \frac{e^2}{k} } \mathcal{E}^2 \\
&= – \inv{2} \alpha \mathcal{E}^2
\end{aligned}

## Van der Wall potential

\label{eqn:qmLecture22:140}
H_0 =
H_{0 1} + H_{0 2},

where

\label{eqn:qmLecture22:160}
H_{0 \alpha} = \frac{p_\alpha^2}{2m} – \frac{e^2}{4 \pi \epsilon_0 \Abs{ \Br_\alpha – \BR_\alpha} }, \qquad \alpha = 1,2

The full interaction potential is

\label{eqn:qmLecture22:180}
V =
\frac{e^2}{4 \pi \epsilon_0} \lr{
\inv{\Abs{\BR_1 – \BR_2}}
+
\inv{\Abs{\Br_1 – \Br_2}}

\inv{\Abs{\Br_1 – \BR_2}}

\inv{\Abs{\Br_2 – \BR_1}}
}

Let

\label{eqn:qmLecture22:200}
\Bx_\alpha = \Br_\alpha – \BR_\alpha,

\label{eqn:qmLecture22:220}
\BR = \BR_1 – \BR_2,

as sketched in fig. 1.

fig. 1. Two atom interaction.

\label{eqn:qmLecture22:240}
H_{0 \alpha}
=
\frac{\Bp^2}{2m}
-\frac{e^2}{4 \pi \epsilon_0 \Abs{\Bx_\alpha}}

which allows the total interaction potential to be written
\label{eqn:qmLecture22:260}
V =
\frac{e^2}{4 \pi \epsilon_0 R}
\lr{
1
+
\frac{R}{\Abs{\Bx_1 – \Bx_2 + \BR}}

\frac{R}{\Abs{\Bx_1 + \BR}}

\frac{R}{\Abs{-\Bx_2 + \BR}}
}

For $$R \gg x_1, x_2$$, this interaction potential, after a multipole expansion, is approximately

\label{eqn:qmLecture22:280}
V =
\frac{e^2}{4 \pi \epsilon_0} \lr{
\frac{\Bx_1 \cdot \Bx_2}{\Abs{\BR}^3}
-3 \frac{
(\Bx_1 \cdot \BR)
(\Bx_2 \cdot \BR)
}{\Abs{\BR}^5}
}

### 1. $$O(\lambda)$$

.

With

\label{eqn:qmLecture22:300}
\psi_0 = \ket{ 1s, 1s }

\label{eqn:qmLecture22:320}
\Delta E^{(1)} = \bra{\psi_0} V \ket{\psi_0}

The two particle wave functions are of the form

\label{eqn:qmLecture22:340}
\braket{ \Bx_1, \Bx_2 }{\psi_0} =
\psi_{1s}(\Bx_1)
\psi_{1s}(\Bx_2),

so braket integrals must be evaluated over a six-fold space. Recall that

\label{eqn:qmLecture22:740}
\psi_{1s} = \inv{\sqrt{\pi} a_0^{3/2} } e^{-r/a_0},

so

\label{eqn:qmLecture22:760}
\bra{\psi_{1s}} x_i \ket{\psi_{1s}}
\propto
\int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi x_i

where
\label{eqn:qmLecture22:780}
x_i \in \setlr{ r \sin\theta \cos\phi, r \sin\theta \sin\phi, r \cos\theta }.

The $$x, y$$ integrals are zero because of the $$\phi$$ integral, and the $$z$$ integral is proportional to $$\int_0^\pi \sin(2 \theta) d\theta$$, which is also zero. This leads to zero averages

\label{eqn:qmLecture22:360}
\expectation{\Bx_1} = 0 = \expectation{\Bx_2}

so

\label{eqn:qmLecture22:380}
\Delta E^{(1)} = 0.

### 2. $$O(\lambda^2)$$

.

\label{eqn:qmLecture22:400}
\begin{aligned}
\Delta E^{(2)}
&= \sum_{n \ne 0} \frac{ \Abs{ \bra{\psi_n } V \ket{\psi_0} }^2 }{E_0 – E_n} \\
&= \sum_{n \ne 0} \frac{ \bra{\psi_0 } V \ket{\psi_n} \bra{\psi_n } V \ket{\psi_0} }{E_0 – E_n}.
\end{aligned}

This is a sum over all excited states. We expect that this will be of the form

\label{eqn:qmLecture22:420}
\Delta E^{(2)} = – \lr{ \frac{e^2}{4 \pi \epsilon_0} }^2 \frac{C_6}{R^6}

$$\Bx_1$$ and $$\Bx_2$$ are dipole operators. The first time this has a non-zero expectation is when we go from the 1s to the 2p states (both 1s and 2s states are spherically symmetric).

Noting that $$E_n = -e^2/2 n^2 a_0$$, we can compute a minimum bound for the energy denominator

\label{eqn:qmLecture22:440}
\begin{aligned}
\lr{E_n – E_0}^{\mathrm{min}}
&= 2 \lr{ E_{2p} – E_{1s} } \\
&= 2 E_{1s} \lr{ \inv{4} – 1 } \\
&= 2 \frac{3}{4} \Abs{E_{1s}} \\
&= \frac{3}{2} \Abs{E_{1s}}.
\end{aligned}

Note that the factor of two above comes from summing over the energies for both electrons. This gives us

\label{eqn:qmLecture22:460}
C_6
=
\frac{3}{2} \Abs{E_{1s}}
\bra{\psi_0 } \tilde{V} \ket{\psi_0},

where

\label{eqn:qmLecture22:480}
\tilde{V} =
\lr{
\Bx_1 \cdot \Bx_2
-3
(\Bx_1 \cdot \Rcap)
(\Bx_2 \cdot \Rcap)
}

\label{eqn:qmLecture22:500}
\Delta E^{(2)}_n
= \sum_{m \ne n} \frac{ \Abs{ \bra{\psi_n } V \ket{\psi_0} }^2 }{E_0 – E_n}

If $$\bra{\psi_n} V \ket{\psi_m} \propto \delta_{n m}$$ then it’s okay.
In general the we can’t expect the matrix element will be anything but fully populated, say

\label{eqn:qmLecture22:520}
V =
\begin{bmatrix}
V_{11} & V_{12} & V_{13} & V_{14} \\
V_{21} & V_{22} & V_{23} & V_{24} \\
V_{31} & V_{32} & V_{33} & V_{34} \\
V_{41} & V_{42} & V_{43} & V_{44} \\
\end{bmatrix},

If we choose a basis so that

\label{eqn:qmLecture22:540}
V =
\begin{bmatrix}
V_{11} & & & \\
& V_{22} & & \\
& & V_{33} & \\
& & & V_{44} \\
\end{bmatrix}.

When this is the case, we have no mixing of elements in the sum of \ref{eqn:qmLecture22:500}

### Degeneracy in the Stark effect

\label{eqn:qmLecture22:560}
H = H_0 + e \mathcal{E} z,

where

\label{eqn:qmLecture22:580}
H_0 = \frac{\Bp^2}{2m} – \frac{e}{4 \pi \epsilon_0} \inv{\Abs{\Bx}}

Consider the states $$2s, 2 p_x, 2p_y, 2p_z$$, for which $$E_n^{(0)} \equiv E_{2 s}$$, as sketched in fig. 2.

fig. 2. 2s 2p degeneracy.

Because of spherical symmetry

\label{eqn:qmLecture22:600}
\begin{aligned}
\bra{2 s} e \mathcal{E} z \ket{ 2 s} &= 0 \\
\bra{2 p_x} e \mathcal{E} z \ket{ 2 p_x} &= 0 \\
\bra{2 p_y} e \mathcal{E} z \ket{ 2 p_y} &= 0 \\
\bra{2 p_z} e \mathcal{E} z \ket{ 2 p_z} &= 0 \\
\end{aligned}

Looking at odd and even properties, it turns out that the only off-diagonal matrix element is

\label{eqn:qmLecture22:620}
\bra{2 s} e \mathcal{E} z \ket{ 2 p_z } = V_1 = -3 e \mathcal{E} a_0.

With a $$\setlr{ 2s, 2p_x, 2p_y, 2p_z }$$ basis the potential matrix is

\label{eqn:qmLecture22:640}
\begin{bmatrix}
0 & 0 & 0 & V_1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
V_1^\conj & 0 & 0 & 0 \\
\end{bmatrix}

\label{eqn:qmLecture22:660}
\begin{bmatrix}
0 & -\Abs{V_1} \\
-\Abs{V_1} & 0 \\
\end{bmatrix}

implies that the energy splitting goes as

\label{eqn:qmLecture22:680}
E_{2s} \rightarrow
E_{2s} \pm \Abs{V_1},

as sketched in fig. 3.

fig. 3. Stark effect energy level splitting.

The diagonalizing states corresponding to eigenvalues $$\pm 3 a_0 \mathcal{E}$$, are $$(\ket{2s} \mp \ket{2p_z})/\sqrt{2}$$.

The matrix element above is calculated explicitly in lecture22Integrals.nb.

The degeneracy that is left unsplit here, and has to be accounted for should we attempt higher order perturbation calculations.

### Appendix. Multipole expansion

Noting that

\label{eqn:qmLecture22:700}
\begin{aligned}
\lr{1 + \epsilon}^{-1/2}
&=
1 -\inv{2} \epsilon -\inv{2}\lr{\frac{-3}{2}}\inv{2!} \epsilon^2 \\
&=
1 -\inv{2} \epsilon + \frac{3}{8} \epsilon^2,
\end{aligned}

we have

\label{eqn:qmLecture22:720}
\begin{aligned}
\frac{R}{\Abs{\Bepsilon + \BR}}
&=
\frac{1}{\Abs{\frac{\Bepsilon}{R} + \Rcap}} \\
&=
\lr{ 1 + 2 \frac{\Bepsilon}{R} \cdot \Rcap + \lr{\frac{\Bepsilon}{R}}^2 }^{-1/2} \\
&=
1 – \frac{\Bepsilon}{R} \cdot \Rcap -\inv{2} \lr{\frac{\Bepsilon}{R}}^2
+ \frac{3}{8}
\lr{ 2 \frac{\Bepsilon}{R} \cdot \Rcap + \lr{\frac{\Bepsilon}{R}}^2 }^2 \\
&=
1 – \frac{\Bepsilon}{R} \cdot \Rcap -\inv{2} \lr{\frac{\Bepsilon}{R}}^2
+ \frac{3}{8}
\lr{ 4 \lr{ \frac{\Bepsilon}{R} \cdot \Rcap}^2 + \lr{\frac{\Bepsilon}{R}}^4
+ 4 \frac{\Bepsilon}{R} \cdot \Rcap \lr{\frac{\Bepsilon}{R}}^2
} \\
&\approx
1 – \frac{\Bepsilon}{R} \cdot \Rcap -\inv{2} \lr{\frac{\Bepsilon}{R}}^2
+ \frac{3}{2}
\lr{ \frac{\Bepsilon}{R} \cdot \Rcap}^2 .
\end{aligned}

Inserting the values from the brackets of \ref{eqn:qmLecture22:260} we have

\label{eqn:qmLecture22:800}
\begin{aligned}
1
+
\frac{R}{\Abs{\Bx_1 – \Bx_2 + \BR}}
&-
\frac{R}{\Abs{\Bx_1 + \BR}}

\frac{R}{\Abs{-\Bx_2 + \BR}} \\
&=
– \frac{\lr{ \Bx_1 – \Bx_2 }}{R} \cdot \Rcap -\inv{2} \lr{\frac{\lr{ \Bx_1 – \Bx_2 }}{R}}^2
+ \frac{3}{2}
\lr{ \frac{\lr{ \Bx_1 – \Bx_2 }}{R} \cdot \Rcap}^2 \\
&\quad + \frac{\Bx_1}{R} \cdot \Rcap +\inv{2} \lr{\frac{\Bx_1}{R}}^2
– \frac{3}{2}
\lr{ \frac{\Bx_1}{R} \cdot \Rcap}^2 \\
&\quad – \frac{\Bx_2}{R} \cdot \Rcap +\inv{2} \lr{\frac{\Bx_2}{R}}^2
– \frac{3}{2}
\lr{ \frac{\Bx_2}{R} \cdot \Rcap}^2 \\
&=
\frac{\Bx_1}{R} \cdot \frac{\Bx_2 }{R}
+ \frac{3}{2}
\lr{ \frac{\lr{ \Bx_1 – \Bx_2 }}{R} \cdot \Rcap}^2 \\
– \frac{3}{2}
\lr{ \frac{\Bx_1}{R} \cdot \Rcap}^2 \\
– \frac{3}{2}
\lr{ \frac{\Bx_2}{R} \cdot \Rcap}^2 \\
&=
\frac{\Bx_1}{R} \cdot \frac{\Bx_2 }{R}
– 3 \frac{\Bx_1}{R} \cdot \Rcap \frac{\Bx_2}{R} \cdot \Rcap.
\end{aligned}

This proves \ref{eqn:qmLecture22:280}.

# 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.