Physics Problems & Solutions: Perturbation Theory

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Quantum Mechanics - Perturbation Theory

The raising and lowering operators for the quantum harmonic oscillator satisfy

$\begin{aligned} a^\dagger |n\rangle&= \sqrt{(n+1)}|n+1\rangle\\ a |n\rangle &= \sqrt{n}|n-1\rangle \end{aligned}$

for energy eigenstates $|n\rangle$ with energy E_n

. Which of the following gives the first-order shift in the n=2

energy level due to the perturbation

$\Delta H=V(a+a^\dagger )^2$

where

is constant?

A.

C. $\sqrt 2 V$
D. $2\sqrt 2 V$
E. 5 V

(GR0177 #94)

Solution:

The energy

$\Delta E=\langle n| \Delta H |n\rangle$

with n=2

and $\Delta H=V(a+a^\dagger )^2$

$\begin{aligned} \Delta E&=\langle 2| V(a+a^\dagger )^2 |2\rangle\\ &=V\langle 2| (aa+a^\dagger a^\dagger + aa^\dagger+a^\dagger a|2\rangle \\ &=V\langle 2|aa|2\rangle+V\langle 2|a^\dagger a^\dagger|2\rangle+V\langle 2|aa^\dagger|2\rangle+V\langle 2|a^\dagger a|2\rangle \end{aligned}$

Raising and lowering operators:

$\begin{aligned} a^\dagger |1\rangle &= \sqrt{1+1}|1+1\rangle=\sqrt{2}|2\rangle\\ a^\dagger |2\rangle &= \sqrt{2+1}|2+1\rangle=\sqrt{3}|3\rangle\\ a^\dagger |3\rangle &= \sqrt{3+1}|3+1\rangle=\sqrt{4}|4\rangle\\ \end{aligned}$

$\begin{aligned} a |1\rangle &= \sqrt{1}|1-1\rangle=\sqrt{1}|0\rangle\\ a |2\rangle &= \sqrt{2}|2-1\rangle=\sqrt{2}|1\rangle\\ a |3\rangle &= \sqrt{3}|3-1\rangle=\sqrt{3}|2\rangle\\ \end{aligned}$

$\langle 2|aa|2\rangle = \langle 2|a\sqrt{2}|1\rangle=\sqrt{2}\langle 2|a|1\rangle=\sqrt{2}\langle 2|\sqrt{1}|0\rangle= \sqrt{2}\sqrt{1}\langle 2|0\rangle=0$

$\langle 2|a^\dagger a^\dagger|2\rangle = \langle 2|a^\dagger\sqrt{3}|3\rangle=\sqrt{3}\langle 2|a^\dagger|3\rangle=\sqrt{3}\langle 2|\sqrt{4}|4\rangle= \sqrt{3}\sqrt{4}\langle 2|4\rangle=0$

$\langle 2|aa^\dagger|2\rangle = \langle 2|a\sqrt{3}|3\rangle=\sqrt{3}\langle 2|a|3\rangle=\sqrt{3}\langle 2|\sqrt{3}|2\rangle= \sqrt{3}\sqrt{3}\langle 2|2\rangle=3$

$\langle 2|a^\dagger a|2\rangle = \langle 2|a^\dagger\sqrt{2}|1\rangle=\sqrt{2}\langle 2|a^\dagger|1\rangle=\sqrt{2}\langle 2|\sqrt{2}|2\rangle= \sqrt{2}\sqrt{2}\langle 2|2\rangle=2$

$\Delta E=3V+2V=5V$

Answer: E

Quantum Mechanics – Perturbation Theory

A particle of mass M is in an infinitely deep square well potential

where

for $-a \leq x \leq a$ , and
$V=\infty$ for x < -a, a < x

.

A very small perturbing potential

' is superimposed on

such that

$V'=\epsilon \left( \frac{a}{2}-\left| x \right| \right)$ for $-\frac{a}{2} \leq x \leq \frac{a}{2}$ , and

for $x < -\frac{a}{2} , \hspace{5mm}\frac{a}{2} < x$ .

If $\psi_0,\psi_1,\psi_2,\psi_3,...$ are the energy eigenfunctions for a particle in the infinitely deep square well potential, with $\psi_0$ being the ground state, which of the following statements is correct about the eigenfunction $\psi_0'$ of a particle in the perturbed potential

?

A.   $\psi_0' = a_{00} \psi_0,$    $a_{00} \neq 0$

B.   $\psi_0'= \sum_{n=0}^\infty a_{0n} \psi_n \hspace{5mm}$ with    $a_{0n}=0 \hspace{5mm}$ for all odd values of n.

C.     $\psi_0'= \sum_{n=0}^\infty a_{0n} \psi_n \hspace{5mm}$ with    $a_{0n}=0 \hspace{5mm}$ for all even values of n.

D.    $\psi_0'= \sum_{n=0}^\infty a_{0n} \psi_n \hspace{5mm}$ with $a_{0n}=0 \hspace{5mm}$ for all values of n.

E.   None of the above

(GR8677 #96)

Solution:

The eigenfunctions of the unperturbed infinite deep well form a complete set (or complete basis).

This means that any function can be represented by the old unperturbed infinite deep well.

Thus, the solution to the perturbed eigenfunction should look like .

$\psi_0' = \sum_{n=0}^\infty a_{0n} \psi_n$

Also, since the perturbed potential is also symmetrical with respect to the origin (as the original unperturbed potential was, too), one knows that all the odd terms should go to 0.

Answer: B

Quantum Mechanics - Perturbation Theory

In perturbation theory, what is the first order correction to the energy of a hydrogen atom (Bohr radius a₀) in its ground state due to the presence of a static electric field E?

A. 0
B. eEa₀
C. 3eEa₀
D. 8e²Ea₀³/ 3
E. 8e²E²a₀³/ 3

(GR9277 #99)

Solution

Answer: A