Quantom Mechanics Chp7

QMHW Chapter 7

7.3

a

For the infinite square wall, the wave function is:
[\psi_n(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi }{a} x\right)]
With energy:
[E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}]
Given two identical Bosons, we have:
Ground state :
[ \psi {gs}(x_1,x_2) = \psi{gs}(x_1)\psi_{gs}(x_2) = \frac{2}{a} \sin\left(\frac{\pi }{a} x_1\right)\sin\left(\frac{\pi }{a} x_2\right)]
And engergy:
[E_{gs} = 2E_{gs} =\frac{\pi^2\hbar^2}{ma^2} ]
According to Example 5.1, we have
[
\Psi_{\text{1st}}(x_1, x_2) = \frac{1}{\sqrt{2}} \left[ \psi_1(x_1)\psi_2(x_2) + \psi_2(x_1)\psi_1(x_2) \right]
\
= \frac{\sqrt 2}{a} \left[ \sin\left(\frac{\pi x_1}{a}\right) \sin\left(\frac{2 \pi x_2}{a}\right) + \sin\left(\frac{2 \pi x_1}{a}\right) \sin\left(\frac{\pi x_2}{a}\right) \right]
]

and

[E_{1st} = E_1+E_2 = \frac{5\pi^2\hbar^2}{2ma^2}]

b

[\begin{aligned}
E_1^1 &= \bra{\psi_1^0} H’ \ket{\psi_1^0} \
&= -a V_0 (2/a)^2 \int \int \sin^2(\pi x_1/a)\sin^2(\pi x_2 / a) \delta (x_1 - x_2) d x_1 dx_2 \
&= -\frac 3 2 V_0\ \
E_2^1 &= \bra{\psi_2^0} H’ \ket{\psi_2^0}\
& = -a V_0 (2/a)^2 \int\int \left[ \sin\left(\frac{\pi x_1}{a}\right) \sin\left(\frac{2 \pi x_2}{a}\right) + \sin\left(\frac{2 \pi x_1}{a}\right) \sin\left(\frac{\pi x_2}{a}\right) \right]^2 \delta (x_1 - x_2) d x_1 dx_2\
&= -2V_0
\end{aligned} ]

7.8

a

[\braket{\psi_{+}^0|\psi_-^0} = \braket{\alpha_+\psi_a^0+ \beta_+ \psi_b^0|\alpha_-\psi_a^0+ \beta_- \psi_b^0} = \alpha_+^\alpha_- + \beta_+^\beta_-]
Given [\bold W \begin{bmatrix}\alpha\\beta\end{bmatrix} = E^1\begin{bmatrix}\alpha\\beta\end{bmatrix} ]
where $\bold W$ is hermitan, its distinct eigenvectors are orthogonal, hense
[ \braket{(\alpha_+, \beta_+)|(\alpha_-, \beta_-) } =0]

b

[ \braket{\psi_+^0|H’|\psi_-^0} = \begin{bmatrix}\alpha_+,\beta_+\end{bmatrix}’ \bold{W}\begin{bmatrix}\alpha_-\\beta_- \end{bmatrix} = E_-\braket{(\alpha_+, \beta_+)|(\alpha_-, \beta_-) } =0]

c

we prove for $+$
[\braket{\psi_+^0|H’|\psi_+^0}=\begin{bmatrix}\alpha_+,\beta_+\end{bmatrix}’ \bold{W}\begin{bmatrix}\alpha_+\\beta_+ \end{bmatrix} = E_+^1 \begin{bmatrix}\alpha_+,\beta_+\end{bmatrix}’ \begin{bmatrix}\alpha_+\\beta_+ \end{bmatrix} = E_+^1]

7.11

The wave function for the 3-dimension infinite square wall is:
[\psi^0(x,y,z) = \left(\frac{2}{a}\right)^{3/2} \sin(\frac{\pi }{a} x)\sin(\frac{\pi }{a} y)\sin(\frac{\pi }{a} z)]
The ground state energy after perturbation is:
[\begin{aligned}E_{gs} &= \braket{\psi^0|H’|\psi^0}\
&=a^3 V_0 \left(\frac{2}{a}\right)^3\int\int\int [\sin(\frac{\pi }{a} x)\sin(\frac{\pi }{a} y)\sin(\frac{\pi }{a} z)]^2 \delta(x-a/4)\delta(y-a/2)\delta(z-3a/4)dxdydz\&=2V_0
\end{aligned}]
The first excited states can be caculated from the eigenvalues of $\bold W$, where $\bold W$ has the formula:
[W_{ij} = \braket{\psi_i^0|H’|\psi_j^0}]
[\bold{W} = \begin{bmatrix} 4V_0 &0 & -4V_0\ 0&0&0\ -4V_0 &0 &4V_0\end{bmatrix}]
Let $\det(\lambda I - \bold W) = 0$, we get $\lambda = 0,0,8V_0$. Hence the correction energies are $0,0,8V_0$

7.41

a

First check that $\psi(x+L) = \psi(x)$
[\exp(\frac{i2\pi n (x+L)}L) =\exp(\frac{i2\pi n x}L + 2n\pi i) = \exp(\frac{i2\pi n x}L) ]
Also check that total probability is $1$
[\braket{\psi_n^0|\psi_n^0} =1]

Check the Sch$\"{o}$dinger equation $ H^0\psi_n^0(x) = E^0 \psi_n^0(x)$ and it does satisfy with energy
[E^0 = \frac{2\pi^2\hbar^2n}{mL^2}]

The energies for $n\geq 1$ can be calculates with:

[E_n=\braket{\psi_n^0|H^0|\psi_n^0} = \frac{2\hbar^2 \pi^2 n^2}{mL^2}]

Notice that for $+n$ and $-n$ they have the same energy correction, therefore for $n=\pm 1,\pm 2 ,\ldots$ are first order degenerate.

b

[\begin{aligned}H’{mn} &= \braket{\psi_m^0|H’|\psi_n^0}\
&= \frac{\hbar^2}{2mL^3} \oint{L}\exp(\frac{2\pi x}{L}(n-m)i) dx\
&=\begin{cases}\frac{\hbar^2}{2mL^2} & n=m\0&n\neq m\end{cases}
\end{aligned}]

c

[\bold W = \frac{\hbar^2}{2mL^2} I]
It can be seen that the eigenvalue is $E^1 = \frac{\hbar^2}{2mL^2}$

d

Using the formula from Problem 7.40 :
[[\bold W^{(2)}]{ij} = \sum{m\neq i,j}\frac{\braket{\psi_i^0|H’|\psi_m^0}\braket{\psi_m^0|H’|\psi_j^0}}{E^0 - E_m^0}]
we have:
[\bold W^{(2)} = \frac{\hbar^2}{2mL^2}\begin{bmatrix}\frac{1}{4\pi^2-1} &\&\frac{1}{-4\pi^2-1}\end{bmatrix}]
It has two different eigenvalues:

E_{+1}^{2} = \frac{\hbar^2}{(2mL^2)(4\pi^2-1) }\quad E_{-1}^{2} = \frac{\hbar^2}{(2mL^2)(-4\pi^2-1) }$$suggesting that degeneracy nolonger holds. The eigenvectors are $(1,0)^T$and $(0,1)^T$, this implies that the given wave functions are already "good". ## e Notice that $E_{+n}^2 = E_{+1}^2$ and $E_{-n}^2 = E_{-1}^2$ , so for all $n$ we have the second order correction: \[E_n = E_n^0 +E_n^1 +E_n^2 = \frac{4\pi^2\hbar^2+\hbar^2}{2mL^2} + \frac{\hbar^2}{2mL^2}\frac{1}{4\pi^2 \mathrm{sgn}(n) -1}\]