Month: December 2015

Harmonic oscillator with energy shift

December 5, 2015 phy1520 , ,

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Q: [1] pr 5.1

Given a perturbed 1D SHO Hamiltonian

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:20}
H = \inv{2m} p^2 + \inv{2} m \omega^2 x^2 + \lambda b x,
\end{equation}

calculate the first non-zero perturbation to the ground state energy. Then solve for that energy directly and compare.

A:

The first order energy shift is seen to be zero

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:40}
\begin{aligned}
\Delta_0^{(0)}
&= V_{00} \\
&= \bra{0} b x \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \bra{0} a + a^\dagger \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \braket{0}{1} \\
&= 0.
\end{aligned}
\end{equation}

The first order perturbation to the ground state is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:60}
\begin{aligned}
\ket{0^{(1)}}
&= \sum_{m \ne 0} \frac{ \ket{m} \bra{m} b x \ket{0} }{ \Hbar \omega/2 – \Hbar
\omega (m – 1/2) } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \sum_{m \ne 0} \frac{ \ket{m}
\braket{m}{1} }{ m } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1}.
\end{aligned}
\end{equation}

The second order ground state energy perturbation is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:80}
\begin{aligned}
\Delta_0^{(2)}
&=
\bra{0} b x \ket{0^{(1)}} \\
&=
\frac{b x_0}{\sqrt{2}} \bra{0} a + a^\dagger \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1} } \\
&=
\frac{b x_0}{\sqrt{2}} \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} } \\
&=
-\frac{b^2 x_0^2}{ 2 \Hbar \omega } \\
&=
-\frac{b^2 }{ 2 \Hbar \omega } \frac{\Hbar}{m \omega} \\
&=
-\frac{b^2 }{ 2 m \omega^2 },
\end{aligned}
\end{equation}

so the total energy perturbation up to second order is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:100}
\Delta_0 = -\lambda^2 \frac{b^2 }{ 2 m \omega^2 }.
\end{equation}

To compare to the exact result, rewrite the Hamiltonian as

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:120}
\begin{aligned}
H
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x^2 + \frac{2 \lambda b x}{m \omega^2} } \\
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x + \frac{\lambda b }{m \omega^2} }^2 – \inv{2} m \omega^2 \lr{ \frac{\lambda b }{m \omega^2} }^2.
\end{aligned}
\end{equation}

The Hamiltonian is subject to a constant energy shift

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:140}
\begin{aligned}
\Delta E
&=
– \inv{2} m \omega^2 \frac{\lambda^2 b^2 }{m^2 \omega^4} \\
&=
– \frac{\lambda^2 b^2 }{2 m \omega^2}.
\end{aligned}
\end{equation}

This is an exact match with the second order perturbation result of \ref{eqn:harmonicOscillatorEnergyShiftPertubation:100}.

References

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

Energy estimate for an absolute value potential

December 4, 2015 phy1520 , , ,

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Here’s a simple problem, a lot like the problem set 6 variational calculation.

Q: [1] 5.21

Estimate the lowest eigenvalue \( \lambda \) of the differential equation

\begin{equation}\label{eqn:absolutePotentialVariation:20}
\frac{d^2}{dx^2}\psi + \lr{ \lambda – \Abs{x} } \psi = 0.
\end{equation}

Using \( \alpha \) variation with the trial function

\begin{equation}\label{eqn:absolutePotentialVariation:40}
\psi =
\left\{
\begin{array}{l l}
c(\alpha – \Abs{x}) & \quad \mbox{\(\Abs{x} < \alpha \) } \\ 0 & \quad \mbox{\(\Abs{x} > \alpha \) }
\end{array}
\right.
\end{equation}

A:

First rewrite the differential equation in a Hamiltonian like fashion

\begin{equation}\label{eqn:absolutePotentialVariation:60}
H \psi = -\frac{d^2}{dx^2}\psi + \Abs{x} \psi = \lambda \psi.
\end{equation}

We need the derivatives of the trial distribution. The first derivative is

\begin{equation}\label{eqn:absolutePotentialVariation:80}
\begin{aligned}
\frac{d}{dx} \psi
&=
-c \frac{d}{dx} \Abs{x} \\
&=
-c \frac{d}{dx} \lr{ x \theta(x) – x \theta(-x) } \\
&=
-c \lr{
\theta(x) – \theta(-x)
+
x \delta(x) + x \delta(-x)
} \\
&=
-c \lr{
\theta(x) – \theta(-x)
+
2 x \delta(x)
}.
\end{aligned}
\end{equation}

The second derivative is
\begin{equation}\label{eqn:absolutePotentialVariation:100}
\begin{aligned}
\frac{d^2}{dx^2} \psi
&=
-c \frac{d}{dx} \lr{
\theta(x) – \theta(-x)
+
2 x \delta(x)
} \\
&=
-c \lr{
\delta(x) + \delta(-x)
+
2 \delta(x)
+
2 x \delta'(x)
} \\
&=
-c \lr{
4 \delta(x)
+
2 x \frac{-\delta(x) }{x}
} \\
&=
-2 c \delta(x).
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:absolutePotentialVariation:120}
H \psi = -2 c \delta(x) + \Abs{x} c \lr{ \alpha – \Abs{x} }.
\end{equation}

We are now set to compute some of the inner products. The normalization is the simplest

\begin{equation}\label{eqn:absolutePotentialVariation:140}
\begin{aligned}
\braket{\psi}{\psi}
&= c^2 \int_{-\alpha}^\alpha ( \alpha – \Abs{x} )^2 dx \\
&= 2 c^2 \int_{0}^\alpha ( x – \alpha )^2 dx \\
&= 2 c^2 \int_{-\alpha}^0 u^2 du \\
&= 2 c^2 \lr{ -\frac{(-\alpha)^3}{3} } \\
&= \frac{2}{3} c^2 \alpha^3.
\end{aligned}
\end{equation}

For the energy
\begin{equation}\label{eqn:absolutePotentialVariation:160}
\begin{aligned}
\braket{\psi}{H \psi}
&=
c^2 \int dx \lr{ \alpha – \Abs{x} } \lr{ -2 \delta(x) + \Abs{x} \lr{ \alpha – \Abs{x} } } \\
&=
c^2 \lr{ – 2 \alpha + \int_{-\alpha}^\alpha dx \lr{ \alpha – \Abs{x} }^2 \Abs{x} } \\
&=
c^2 \lr{ – 2 \alpha + 2 \int_{-\alpha}^0 du u^2 \lr{ u + \alpha } } \\
&=
c^2 \lr{ – 2 \alpha + 2 \evalrange{\lr{ \frac{u^4}{4} + \alpha \frac{u^3}{3} }}{-\alpha}{0} } \\
&=
c^2 \lr{ – 2 \alpha – 2 \lr{ \frac{\alpha^4}{4} – \frac{\alpha^4}{3} } } \\
&=
c^2 \lr{ – 2 \alpha + \inv{6} \alpha^4 }.
\end{aligned}
\end{equation}

The energy estimate is

\begin{equation}\label{eqn:absolutePotentialVariation:180}
\begin{aligned}
\overline{{E}}
&=
\frac{\braket{\psi}{H \psi}}{\braket{\psi}{\psi}} \\
&=
\frac{ – 2 \alpha + \inv{6} \alpha^4 }{ \frac{2}{3} \alpha^3} \\
&=
– \frac{3}{\alpha^2} + \inv{4} \alpha.
\end{aligned}
\end{equation}

This has its minimum at
\begin{equation}\label{eqn:absolutePotentialVariation:200}
0 = -\frac{6}{\alpha^3} + \inv{4},
\end{equation}

or
\begin{equation}\label{eqn:absolutePotentialVariation:220}
\alpha = 2 \times 3^{1/3}.
\end{equation}

Back subst into the energy gives

\begin{equation}\label{eqn:absolutePotentialVariation:240}
\begin{aligned}
\overline{{E}}
&=
– \frac{3}{4 \times 3^{2/3}} + \inv{2} 3^{1/3} \\
&= \frac{3^{4/3}}{4} \\
&\approx 1.08.
\end{aligned}
\end{equation}

The problem says the exact answer is 1.019, so the variation gets within 6 %.

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 21: Non-degenerate perturbation. Taught by Prof. Arun Paramekanti

December 4, 2015 phy1520 , , ,

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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 [2] chap. 5 content.

Non-degenerate perturbation theory. Recap.

\begin{equation}\label{eqn:qmLecture21:20}
\ket{n} = \ket{n_0}
+ \lambda \ket{n_1}
+ \lambda^2 \ket{n_2}
+ \lambda^3 \ket{n_3} + \cdots
\end{equation}

and

\begin{equation}\label{eqn:qmLecture21:40}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots
\end{equation}

\begin{equation}\label{eqn:qmLecture21:60}
\begin{aligned}
\Delta_{n_1} &= \bra{n^{(0)}} V \ket{n^{(0)}} \\
\ket{n_0} &= \ket{n^{(0)}}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture21:80}
\begin{aligned}
\Delta_{n_2} &= \sum_{m \ne n} \frac{\Abs{\bra{n^{(0)}} V \ket{m^{(0)}}}^2}{E_n^{(0)} – E_m^{(0)}} \\
\ket{n_1} &= \sum_{m \ne n} \frac{ \ket{m^{(0)}} V_{mn} }{E_n^{(0)} – E_m^{(0)}}
\end{aligned}
\end{equation}

Example: Stark effect

\begin{equation}\label{eqn:qmLecture21:100}
H = H_{\textrm{atom}} + e \mathcal{E} z,
\end{equation}

where \( H_{\textrm{atom}} \) is assumed to be Hydrogen-like with Hamiltonian

\begin{equation}\label{eqn:qmLecture21:120}
H_{\textrm{atom}} = \frac{\BP^2}{2m} – \frac{e^2}{4 \pi \epsilon_0 r},
\end{equation}

and wave functions

\begin{equation}\label{eqn:qmLecture21:140}
\braket{\Br}{\psi_{n l m}} = R_{n l}(r) Y_{lm}( \theta, \phi )
\end{equation}

For the first level correction to the energy

\begin{equation}\label{eqn:qmLecture21:160}
\begin{aligned}
\Delta_1
&= \bra{\psi_{100}} e \mathcal{E} z \ket{ \psi_{100}} \\
&= e \mathcal{E} \int \frac{d\Omega}{4 \pi} \cos \theta \int dr r^2 R_{100}^2(r)
\end{aligned}
\end{equation}

The cosine integral is obliterated, so we have \( \Delta_1 = 0 \).

How about the second order energy correction? That is

\begin{equation}\label{eqn:qmLecture21:180}
\Delta_2 = \sum_{n l m \ne 100} \frac{
\Abs{ \bra{\psi_{100}} e \mathcal{E} z \ket{ n l m }}^2
}{
E_{100}^{(0)} – E_{n l m}
}
\end{equation}

The matrix element in the numerator is the absolute square of

\begin{equation}\label{eqn:qmLecture21:200}
V_{100,nlm}
=
e \mathcal{E} \int d\Omega \inv{\sqrt{ 4 \pi } }
\cos\theta Y_{l m}(\theta, \phi)
\int dr r^3 R_{100}(r) R_{n l}(r).
\end{equation}

For all \( m \ne 0 \), \( Y_{lm} \) includes a \( e^{i m \phi} \) factor, so this cosine integral is zero. For \( m = 0 \), each of the \( Y_{lm} \) functions appears to contain either even or odd powers of cosines. For example:

\begin{equation}\label{eqn:qmLecture21:760}
\begin{aligned}
Y_{00} &= \frac{1}{2 \sqrt{\pi}} \\
Y_{10} &= \frac{1}{2} \sqrt{\frac{3}{\pi }} \cos(t) \\
Y_{20} &= \frac{1}{4} \sqrt{\frac{5}{\pi }} \lr{(3 \cos^2(t)-1} \\
Y_{30} &= \frac{1}{4} \sqrt{\frac{7}{\pi }} \lr{(5 \cos^3(t)-3 \cos(t)} \\
Y_{40} &= \frac{3 \lr{(35 \cos^4(t)-30 \cos^2(t)+3}}{16 \sqrt{\pi }} \\
Y_{50} &= \frac{1}{16} \sqrt{\frac{11}{\pi }} \lr{(63 \cos^5(t)-70 \cos^3(t)+15 \cos(t)} \\
Y_{60} &= \frac{1}{32} \sqrt{\frac{13}{\pi }} \lr{(231 \cos^6(t)-315 \cos^4(t)+105 \cos^2(t)-5} \\
Y_{70} &= \frac{1}{32} \sqrt{\frac{15}{\pi }} \lr{(429 \cos^7(t)-693 \cos^5(t)+315 \cos^3(t)-35 \cos(t)} \\
Y_{80} &= \frac{1}{256} \sqrt{\frac{17}{\pi }} \lr{(6435 \cos^8(t)-12012 \cos^6(t)+6930 \cos^4(t)-1260 \cos^2(t)+35 } \\
\end{aligned}
\end{equation}

This shows that for even \( 2k = l \), the cosine integral is zero

\begin{equation}\label{eqn:qmLecture21:780}
\int_0^\pi \sin\theta \cos\theta \sum_k a_k \cos^{2k}\theta d\theta
=
0,
\end{equation}

since \( \cos^{2k}(\theta) \) is even and \( \sin\theta \cos\theta \) is odd over the same interval. We find zero for \( \int_0^\pi \sin\theta \cos\theta Y_{30}(\theta, \phi) d\theta \), and Mathematica appears to show that the rest of these integrals for \( l > 1 \) are also zero.

FIXME: find the property of the spherical harmonics that can be used to prove that this is true in general for \( l > 1 \).

This leaves

\begin{equation}\label{eqn:qmLecture21:220}
\begin{aligned}
\Delta_2
&= \sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} e \mathcal{E} z \ket{ n 1 0 }}^2
}{
E_{100}^{(0)} – E_{n 1 0}
} \\
&=
-e^2 \mathcal{E}^2
\sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} z \ket{ n 1 0 }}^2
}{
E_{n 1 0}
-E_{100}^{(0)}
}.
\end{aligned}
\end{equation}

This is sometimes written in terms of a polarizability \( \alpha \)

\begin{equation}\label{eqn:qmLecture21:260}
\Delta_2 = -\frac{\mathcal{E}^2}{2} \alpha,
\end{equation}

where

\begin{equation}\label{eqn:qmLecture21:280}
\alpha =
2 e^2
\sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} z \ket{ n 1 0 }}^2
}{
E_{n 1 0}
-E_{100}^{(0)}
}.
\end{equation}

With
\begin{equation}\label{eqn:qmLecture21:840}
\BP = \alpha \boldsymbol{\mathcal{E}},
\end{equation}

the energy change upon turning on the electric field from \( 0 \rightarrow \mathcal{E} \) is simply \( – \BP \cdot d\boldsymbol{\mathcal{E}} \) integrated from \( 0 \rightarrow \mathcal{E} \). Putting \( \BP = \alpha \mathcal{E} \zcap \), we have

\begin{equation}\label{eqn:qmLecture21:400}
\begin{aligned}
– \int_0^\mathcal{E} p_z d\mathcal{E}
&=
– \int_0^\mathcal{E} \alpha \mathcal{E} d\mathcal{E} \\
&=
– \inv{2} \alpha \mathcal{E}^2
\end{aligned}
\end{equation}

leading to an energy change \( – \alpha \mathcal{E}^2/2 \), so we can directly compute \( \expectation{\BP} \) or we can compute change in energy, and both contain information about the polarization factor \( \alpha \).

There is an exact answer to the sum \ref{eqn:qmLecture21:280}, but we aren’t going to try to get it here. Instead let’s look for bounds

\begin{equation}\label{eqn:qmLecture21:240}
\Delta_2^{\mathrm{min}} < \Delta_2 < \Delta_2^{\mathrm{max}}
\end{equation}

\begin{equation}\label{eqn:qmLecture21:320}
\alpha^{\mathrm{min}} = 2 e^2 \frac{
\Abs{ \bra{\psi_{100}} z \ket{\psi_{210}} }^2
}{E_{210}^{(0)} – E_{100}^{(0)}}
\end{equation}

For the hydrogen atom we have

\begin{equation}\label{eqn:qmLecture21:820}
E_n = -\frac{ e^2}{ 2 n^2 a_0 },
\end{equation}

allowing any difference of energy levels to be expressed as a fraction of the ground state energy, such as

\begin{equation}\label{eqn:qmLecture21:340}
E_{210}^{(0)} = \inv{4} E_{100}^{(0)} = \inv{4} \frac{ -\Hbar^2 }{ 2 m a_0^2 }
\end{equation}

So
\begin{equation}\label{eqn:qmLecture21:360}
E_{210}^{(0)} – E_{100}^{(0)} = \frac{3}{4}
\frac{ \Hbar^2 }{ 2 m a_0^2 }
\end{equation}

In the numerator we have

\begin{equation}\label{eqn:qmLecture21:380}
\begin{aligned}
\bra{\psi_{100}} z \ket{\psi_{210}}
&=
\int r^2 d\Omega
\lr{ \inv{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0} } r \cos\theta \lr{
\inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \cos\theta
} \\
&=
(2 \pi)
\inv{\sqrt{\pi}} \inv{4 \sqrt{2 \pi} } a_0
\int_0^\pi d\theta \sin\theta \cos^2\theta
\int_0^\infty \frac{dr}{a_0} \frac{r^4}{a_0^4} e^{-r/a_0 – r/2 a_0} \\
&=
(2 \pi)
\inv{\sqrt{\pi}} \inv{4 \sqrt{2 \pi} } a_0
\lr{ \evalrange{-\frac{u^3}{3}}{1}{-1} }
\int_0^\infty s^4 ds e^{- 3 s/2 } \\
&=
2
\inv{4 \sqrt{2} } a_0
\lr{ \evalrange{-\frac{u^3}{3}}{1}{-1} }
\int_0^\infty s^4 ds e^{- 3 s/2 } \\
&=
\inv{2 \sqrt{2}} \frac{2}{3} a_0 \frac{256}{81} \\
&=
\frac{1}{3 \sqrt{2} } \frac{ 256}{81} a_0
\approx 0.75 a_0.
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture21:420}
\begin{aligned}
\alpha^{\mathrm{min}}
&= \frac{ 2 e^2 (0.75)^2 a_0^2 }{ \frac{3}{4} \frac{\Hbar^2}{2 m a_0^2} } \\
&= \frac{6}{4} \frac{2 m e^2 a_0^4}{ \Hbar^2 } \\
&= 3 \frac{m e^2 a_0^4}{ \Hbar^2 } \\
&= 3 \frac{ 4 \pi \epsilon_0 }{a_0} a_0^4 \\
&\approx 4 \pi \epsilon_0 a_0^3 \times 3.
\end{aligned}
\end{equation}

The factor \( 4 \pi \epsilon_0 a_0^3 \) are the natural units for the polarizability.

There is a neat trick that generalizes to many problems to find the upper bound. Recall that the general polarizability was

\begin{equation}\label{eqn:qmLecture21:440}
\alpha
=
2 e^2
\sum_{nlm \ne 100} \frac{
\Abs{ \bra{100} z \ket{ n l m }}^2
}{
E_{n l m}
-E_{100}^{(0)}
}.
\end{equation}

If we are looking for the upper bound, and replace the denominator by the smallest energy difference that will be encountered, it can be brought out of the sum, for

\begin{equation}\label{eqn:qmLecture21:460}
\alpha^{\mathrm{max}} =
2 e^2
\inv{E_{2 1 0}
-E_{100}^{(0)} }
\sum_{nlm \ne 100}
\bra{100} z \ket{ n l m } \bra{nlm} z \ket{ 100 }
\end{equation}

Because \( \bra{nlm} z \ket{100} = 0 \), the constraint in the sum can be removed, and the identity summation evaluated

\begin{equation}\label{eqn:qmLecture21:480}
\begin{aligned}
\alpha^{\mathrm{max}}
&=
2 e^2
\inv{E_{2 1 0}
-E_{100}^{(0)} }
\sum_{nlm}
\bra{100} z \ket{ n l m } \bra{nlm} z \ket{ 100 } \\
&=
\frac{2 e^2 }{ \frac{3}{4} \frac{\Hbar^2}{ 2 m a_0^2} }
\bra{100} z^2 \ket{ 100 } \\
&=
\frac{16 e^2 m a_0^2 }{ 3 \Hbar^2 } \times a_0^2 \\
&=
4 \pi \epsilon_0 a_0^3 \times \frac{16}{3}.
\end{aligned}
\end{equation}

The bounds are

\begin{equation}\label{eqn:qmLecture21:520}
\boxed{
3 \ge \frac{\alpha}{\alpha^{\mathrm{at}}} < \frac{16}{3},
}
\end{equation}

where

\begin{equation}\label{eqn:qmLecture21:560}
\alpha^{\mathrm{at}} = 4 \pi \epsilon_0 a_0^3.
\end{equation}

The actual value is
\begin{equation}\label{eqn:qmLecture21:580}
\frac{\alpha}{\alpha^{\mathrm{at}}} = \frac{9}{2}.
\end{equation}

Example: Computing the dipole moment

\begin{equation}\label{eqn:qmLecture21:600}
\expectation{P_z}
= \alpha \mathcal{E}
= \bra{\psi_{100}} e z \ket{\psi_{100}}.
\end{equation}

Without any perturbation this is zero. After perturbation, retaining only the terms that are first order in \( \delta \psi_{100} \) we have

\begin{equation}\label{eqn:qmLecture21:620}
\bra{\psi_{100} + \delta \psi_{100}} e z \ket{\psi_{100} + \delta \psi_{100}}
\approx
\bra{\psi_{100}} e z \ket{\delta \psi_{100}}
+
\bra{\delta \psi_{100}} e z \ket{\psi_{100}}.
\end{equation}

Next time: Van der Walls

We will look at two hyrdogenic atomic systems interacting where the pair of nuclei are supposed to be infinitely heavy and stationary. The wave functions each set of atoms are individually known, but we can consider the problem of the interactions of atom 1’s electrons with atom 2’s nucleus and atom 2’s electrons, and also the opposite interactions of atom 2’s electrons with atom 1’s nucleus and its electrons. This leads to a result that is linear in the electric field (unlike the above result, which is called the quadratic Stark effect).

Appendix. Hydrogen wavefunctions

From [3], with the \( a_0 \) factors added in.

\begin{equation}\label{eqn:qmLecture21:660}
\psi_{1 s} = \psi_{100} = \inv{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture21:680}
\psi_{2 s} = \psi_{200} = \inv{4 \sqrt{2 \pi} a_0^{3/2}} \lr{ 2 – \frac{r}{a_0} } e^{-r/2a_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture21:700}
\psi_{2 p_x} = \inv{\sqrt{2}} \lr{ \psi_{2,1,1} – \psi_{2,1,-1} }
= \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \sin\theta\cos\phi
\end{equation}
\begin{equation}\label{eqn:qmLecture21:720}
\psi_{2 p_y} = \frac{i}{\sqrt{2}} \lr{ \psi_{2,1,1} + \psi_{2,1,-1} }
= \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \sin\theta\sin\phi
\end{equation}
\begin{equation}\label{eqn:qmLecture21:740}
\psi_{2 p_z} = \psi_{210} = \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \cos\theta
\end{equation}

I looked to [1] to see where to add in the \( a_0 \) factors.

References

[1] Carl R. Nave. Hydrogen Wavefunctions, 2015. URL http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html. [Online; accessed 03-Dec-2015].

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

[3] Robert Field Troy Van Voorhis. Hydrogen Atom, 2013. URL https://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2013/lecture-notes/MIT5_61F13_Lecture19-20.pdf. [Online; accessed 03-Dec-2015].

PHY1520H Graduate Quantum Mechanics. Lecture 20: Perturbation theory. Taught by Prof. Arun Paramekanti

December 3, 2015 phy1520 , , , , , , , , , ,

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

Disclaimer

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

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] ch. 5 content.

Perturbation theory

Given a \( 2 \times 2 \) Hamiltonian \( H = H_0 + V \), where

\begin{equation}\label{eqn:qmLecture20:20}
H =
\begin{bmatrix}
a & c \\
c^\conj & b
\end{bmatrix}
\end{equation}

which has eigenvalues

\begin{equation}\label{eqn:qmLecture20:40}
\lambda_\pm = \frac{a + b}{2} \pm \sqrt{ \lr{ \frac{a – b}{2}}^2 + \Abs{c}^2 }.
\end{equation}

If \( c = 0 \),

\begin{equation}\label{eqn:qmLecture20:60}
H_0 =
\begin{bmatrix}
a & 0 \\
0 & b
\end{bmatrix},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:80}
V =
\begin{bmatrix}
0 & c \\
c^\conj & 0
\end{bmatrix}.
\end{equation}

Suppose that \( \Abs{c} \ll \Abs{a – b} \), then

\begin{equation}\label{eqn:qmLecture20:100}
\lambda_\pm \approx \frac{a + b}{2} \pm \Abs{ \frac{a – b}{2} } \lr{ 1 + 2 \frac{\Abs{c}^2}{\Abs{a – b}^2} }.
\end{equation}

If \( a > b \), then

\begin{equation}\label{eqn:qmLecture20:120}
\lambda_\pm \approx \frac{a + b}{2} \pm \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} }.
\end{equation}

\begin{equation}\label{eqn:qmLecture20:140}
\begin{aligned}
\lambda_{+}
&= \frac{a + b}{2} + \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&= a + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= a + \frac{\Abs{c}^2}{a – b},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:qmLecture20:680}
\begin{aligned}
\lambda_{-}
&= \frac{a + b}{2} – \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&=
b + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= b + \frac{\Abs{c}^2}{a – b}.
\end{aligned}
\end{equation}

This adiabatic evolution displays a “level repulsion”, quadradic in \( \Abs{c} \) as sketched in fig. 1, and is described as a non-degenerate perbutation.

fig. 1.  Adiabatic (non-degenerate) perturbation

fig. 1. Adiabatic (non-degenerate) perturbation

If \( \Abs{c} \gg \Abs{a -b} \), then

\begin{equation}\label{eqn:qmLecture20:160}
\begin{aligned}
\lambda_\pm
&= \frac{a + b}{2} \pm \Abs{c} \sqrt{ 1 + \inv{\Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&\approx \frac{a + b}{2} \pm \Abs{c} \lr{ 1 + \inv{2 \Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&= \frac{a + b}{2} \pm \Abs{c} \pm \frac{\lr{a – b}^2}{8 \Abs{c}}.
\end{aligned}
\end{equation}

Here we loose the adiabaticity, and have “level repulsion” that is linear in \( \Abs{c} \), as sketched in fig. 2. We no longer have the sign of \( a – b \) in the expansion. This is described as a degenerate perbutation.

fig. 2.  Degenerate perbutation

fig. 2. Degenerate perbutation

General non-degenerate perturbation

Given an unperturbed system with solutions of the form

\begin{equation}\label{eqn:qmLecture20:180}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}},
\end{equation}

we want to solve the perturbed Hamiltonian equation

\begin{equation}\label{eqn:qmLecture20:200}
\lr{ H_0 + \lambda V } \ket{ n } = \lr{ E_n^{(0)} + \Delta n } \ket{n}.
\end{equation}

Here \( \Delta n \) is an energy shift as that goes to zero as \( \lambda \rightarrow 0 \). We can write this as

\begin{equation}\label{eqn:qmLecture20:220}
\lr{ E_n^{(0)} – H_0 } \ket{ n } = \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

We are hoping to iterate with application of the inverse to an initial estimate of \( \ket{n} \)

\begin{equation}\label{eqn:qmLecture20:240}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

This gets us into trouble if \( \lambda \rightarrow 0 \), which can be fixed by using

\begin{equation}\label{eqn:qmLecture20:260}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

which can be seen to be a solution to \ref{eqn:qmLecture20:220}. We want to ask if

\begin{equation}\label{eqn:qmLecture20:280}
\lr{ \lambda V – \Delta_n } \ket{n} ,
\end{equation}

contains a bit of \( \ket{ n^{(0)} } \)? To determine this act with \( \bra{n^{(0)}} \) on the left

\begin{equation}\label{eqn:qmLecture20:300}
\begin{aligned}
\bra{ n^{(0)} } \lr{ \lambda V – \Delta_n } \ket{n}
&=
\bra{ n^{(0)} } \lr{ E_n^{(0)} – H_0 } \ket{n} \\
&=
\lr{ E_n^{(0)} – E_n^{(0)} } \braket{n^{(0)}}{n} \\
&=
0.
\end{aligned}
\end{equation}

This shows that \( \ket{n} \) is entirely orthogonal to \( \ket{n^{(0)}} \).

Define a projection operator

\begin{equation}\label{eqn:qmLecture20:320}
P_n = \ket{n^{(0)}}\bra{n^{(0)}},
\end{equation}

which has the idempotent property \( P_n^2 = P_n \) that we expect of a projection operator.

Define a rejection operator
\begin{equation}\label{eqn:qmLecture20:340}
\overline{{P}}_n
= 1 –
\ket{n^{(0)}}\bra{n^{(0)}}
= \sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}.
\end{equation}

Because \( \ket{n} \) has no component in the direction \( \ket{n^{(0)}} \), the rejection operator can be inserted much like we normally do with the identity operator, yielding

\begin{equation}\label{eqn:qmLecture20:360}
\ket{n}’ = \lr{ E_n^{(0)} – H_0 }^{-1} \overline{{P}}_n \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

valid for any initial \( \ket{n} \).

Power series perturbation expansion

Instead of iterating, suppose that the unknown state and unknown energy difference operator can be expanded in a \( \lambda \) power series, say

\begin{equation}\label{eqn:qmLecture20:380}
\ket{n}
=
\ket{n_0}
+ \lambda \ket{n_1}
+ \lambda^2 \ket{n_2}
+ \lambda^3 \ket{n_3} + \cdots
\end{equation}

and

\begin{equation}\label{eqn:qmLecture20:400}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots
\end{equation}

We usually interpret functions of operators in terms of power series expansions. In the case of \( \lr{ E_n^{(0)} – H_0 }^{-1} \), we have a concrete interpretation when acting on one of the unpertubed eigenstates

\begin{equation}\label{eqn:qmLecture20:420}
\inv{ E_n^{(0)} – H_0 } \ket{m^{(0)}} =
\inv{ E_n^{(0)} – E_m^0 } \ket{m^{(0)}}.
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture20:440}
\ket{n}
=
\inv{ E_n^{(0)} – H_0 }
\sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}
\lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:460}
\boxed{
\ket{n}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_n } \ket{n}.
}
\end{equation}

From \ref{eqn:qmLecture20:220}, note that

\begin{equation}\label{eqn:qmLecture20:500}
\Delta_n =
\frac{\bra{n^{(0)}} \lambda V \ket{n}}{\braket{n^0}{n}},
\end{equation}

however, we will normalize by setting \( \braket{n^0}{n} = 1 \), so

\begin{equation}\label{eqn:qmLecture20:521}
\boxed{
\Delta_n =
\bra{n^{(0)}} \lambda V \ket{n}.
}
\end{equation}

to \( O(\lambda^0) \)

If all \( \lambda^n, n > 0 \) are zero, then we have

\label{eqn:qmLecture20:780}
\begin{equation}\label{eqn:qmLecture20:740}
\ket{n_0}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ – \Delta_{n_0} } \ket{n_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture20:800}
\Delta_{n_0} \braket{n^{(0)}}{n_0} = 0
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:540}
\begin{aligned}
\ket{n_0} &= \ket{n^{(0)}} \\
\Delta_{n_0} &= 0.
\end{aligned}
\end{equation}

to \( O(\lambda^1) \)

Requiring identity for all \( \lambda^1 \) terms means

\begin{equation}\label{eqn:qmLecture20:760}
\ket{n_1} \lambda
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_{n_1} \lambda } \ket{n_0},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:560}
\ket{n_1}
=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}} \bra{ m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} } \ket{n_0}.
\end{equation}

With the assumption that \( \ket{n^{(0)}} \) is normalized, and with the shorthand

\begin{equation}\label{eqn:qmLecture20:600}
V_{m n} = \bra{ m^{(0)}} V \ket{n^{(0)}},
\end{equation}

that is

\begin{equation}\label{eqn:qmLecture20:580}
\begin{aligned}
\ket{n_1}
&=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}
\\
\Delta_{n_1} &= \bra{ n^{(0)} } V \ket{ n^0} = V_{nn}.
\end{aligned}
\end{equation}

to \( O(\lambda^2) \)

The second order perturbation states are found by selecting only the \( \lambda^2 \) contributions to

\begin{equation}\label{eqn:qmLecture20:820}
\lambda^2 \ket{n_2}
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – (\lambda \Delta_{n_1} + \lambda^2 \Delta_{n_2}) }
\lr{
\ket{n_0}
+ \lambda \ket{n_1}
}.
\end{equation}

Because \( \ket{n_0} = \ket{n^{(0)}} \), the \( \lambda^2 \Delta_{n_2} \) is killed, leaving

\begin{equation}\label{eqn:qmLecture20:840}
\begin{aligned}
\ket{n_2}
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\ket{n_1} \\
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\sum_{l \ne n}
\frac{
\ket{l^{(0)}}
}
{
E_n^{(0)} – E_l^{(0)}
}
V_{l n},
\end{aligned}
\end{equation}

which can be written as

\begin{equation}\label{eqn:qmLecture20:620}
\ket{n_2}
=
\sum_{l,m \ne n}
\ket{m^{(0)}}
\frac{V_{m l} V_{l n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{m \ne n}
\ket{m^{(0)}}
\frac{V_{n n} V_{m n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

For the second energy perturbation we have

\begin{equation}\label{eqn:qmLecture20:860}
\lambda^2 \Delta_{n_2} =
\bra{n^{(0)}} \lambda V \lr{ \lambda \ket{n_1} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:880}
\begin{aligned}
\Delta_{n_2}
&=
\bra{n^{(0)}} V \ket{n_1} \\
&=
\bra{n^{(0)}} V
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}.
\end{aligned}
\end{equation}

That is

\begin{equation}\label{eqn:qmLecture20:900}
\Delta_{n_2}
=
\sum_{m \ne n} \frac{V_{n m} V_{m n} }{E_n^{(0)} – E_m^{(0)}}.
\end{equation}

to \( O(\lambda^3) \)

Similarily, it can be shown that

\begin{equation}\label{eqn:qmLecture20:640}
\Delta_{n_3} =
\sum_{l, m \ne n} \frac{V_{n m} V_{m l} V_{l n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{ m \ne n} \frac{V_{n m} V_{n n} V_{m n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

In general, the energy perturbation is given by

\begin{equation}\label{eqn:qmLecture20:660}
\Delta_n^{(l)} = \bra{n^{(0)}} V \ket{n^{(l-1)}}.
\end{equation}

References

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