4.2 - Separation of variables in Cartesian coordinates means $$\psi(x,y,z)=X(x)Y(y)Z(z).$$

Problem 4.2

part c: The energy of the 14th level is $E_{14}=\frac{27\pi^2\hbar^2}{2ma^2}$. The combinations that give rise to this energy are $\{n_x,n_y,n_z\}=\{1,1,5\},\{1,5,1\},\{5,1,1\},\{3,3,3\}$. What's unique?: All the other energy levels consist of permutations of just one triplet of integers $\{n_x,n_y,n_z\}$. But here, two different triplets of integers have the same energy.

Assignment #9 - Friday, April 5

Read these 4 pages from The Principles of Quantum Mechanics by P.A.M. Dirac. Write: Summarize, with full sentences, but in outline form, the reasons that Dirac gives for a theory beyond classical mechanics. Write a paragraph about the one reason you find most compelling, and why.

Griffiths
Chapter 4: problems 10, 11a, 12b, 13ab, 16, 25, 27

Notes/hints

Assignment #10 -

Griffiths
Chapter 4: 34b, 35, 38, 49 "Bell problems"--Chapter 4: 50*; Chapter 12: 1; Paul's Last problem

Notes/hints

Problem 4.50 - Here's how to figure out the $\hat S_\alpha^{(2)}$ operator.

Selected answers

Problem 4.34b - Show that $\hat S_\pm\ket{00}=0$.

I'll show this for the raising operator. The proof for the lowering operator proceeds along the same lines. Using $\hat S_+= \hat S^{(1)}_+ + \hat S^{(2)}_+$ where $\hat S_+^{(1)}\downarrow^{(1)}=\hbar\uparrow^{(1)}$ and $\hat S_+^{(1)}\uparrow^{(1)}=0$. You get the same thing for the second spin with (1)$\to$(2). $$\begineq \hat S_+\ket{00}&=&(\hat S^{(1)}_+ + \hat S^{(2)}_+)\frac{1}{\sqrt 2} [\uparrow\downarrow - \downarrow\uparrow]\\ &=&\frac{1}{\sqrt 2} \[(\hat S^{(1)}_+ \uparrow)\downarrow - (\hat S^{(1)}_+ \downarrow)\uparrow + \uparrow(\hat S^{(2)}_+\downarrow) - \downarrow(\hat S^{(2)}_+\uparrow)\]\\ &=&\frac{1}{\sqrt 2} \[0 - \hbar \uparrow\uparrow + \hbar\uparrow\uparrow - 0\]\\ &=& 0 \endeq$$

Problem 4.49d - If you measure the $y$-component of spin on the state $\chi=1/3\begincv 1-2i\\2 \endcv$ (in the $S_z$ basis), what is the chance of measuring $+\hbar/2$?

We already found the eigenstates of $\hat S_y\equiv\frac{\hbar}{2} \begincv 0 & -i\\i&0\endcv$. They are: $$ \chi_+^{(y)}=\frac{1}{\sqrt{2}}\begincv -i\\1 \endcv; \ \chi_-^{(y)}=\frac{1}{\sqrt{2}}\begincv i\\1 \endcv\ . $$ with eigenvalues $+\hbar/2$ and $-\hbar/2$.

The state $\chi$ can be written as a superposition of the two $\hat S_y$ eigenstates: $$\chi = c_+\chi_+^{(y)} + c_-\chi_-^{(y)}.$$ The probability of measuring $S_y=+\hbar/2$ is equal to $|c_+|^2$, where $$\begineq c_+&=&\innerp{\chi_+^{(y)}}{\chi}=\chi_+^{(y)\dagger}\chi\\ &=&\frac{1}{\sqrt 2}\begincv +i & 1\endcv 1/3\begincv 1-2i\\2 \endcv\\ &=&\frac{1}{\sqrt {18}}\begincv +i & 1\endcv \begincv 1-2i\\2 \endcv\\ c_+&=&\frac{1}{\sqrt {18}}(+i +2+2)=\frac{4+i}{\sqrt {18}}. \endeq $$ So, the probability of measuring $+\hbar/2$ is $|c_+|^2$: $$|c_+|^2=c_+^*c_+=\frac{4-i}{\sqrt {18}}\frac{4+i}{\sqrt {18}}=\frac{16+1}{18}=\frac{17}{18}.$$ We could use the same method to calculate $c_-$. But since there are only two states, and the probabilities must sum to 1, this means $|c_+|^2+ |c_-|^2=1$. So, $|c_-|^2=1-17/18=1/18$. So, the probability of measuring $S_y=-\hbar/2$ is 1/18.

There are two ways to calculate the expectation value: We could do it this way: $$\langle \hat S_y \rangle=\bra{\chi}\frac{\hbar}{2}\begincv 0&-i\\+i&0\endcv\ket{\chi}.$$ Or, since we know the probabilities, we could take the probability weighted average: $$\langle \hat S_y \rangle=\left(+\frac{\hbar}{2}\right)P_+ + \left(-\frac{\hbar}{2}\right)P_- =\frac{17}{18}\frac{\hbar}{2}-\frac{1}{18}\frac{\hbar}{2}=\frac{16}{36}\hbar=\frac{4}{9}\hbar$$ Either approach gives the same answer.

Last problem - Bell's inequality is: $$1+P(\myv b, \myv c)\geq |P(\myv a,\myv b)-P(\myv a, \myv c)|$$ Since the vectors in $P$ are all unit vectors, $$P(\myv v_1,\myv v_2)=-\myv v_1\cdot\myv v_2=-|v_1||v_2|\cos\alpha = -\cos\alpha,$$ where $\alpha$ is the angle between $\myv v_1$ and $\myv v_2$.

1. For the case that $\myv a=\hat z$ and $\myv b=\hat x$: $P(\myv a,\myv b)=0$, $P(\myv a,\myv c)=-\cos\theta$, $P(\myv b,\myv c)=-\cos(\pi/2-\theta)$.
   
2. For the case that $\myv a=\hat z$ and $\myv b=-\hat z$: $P(\myv a,\myv b)=1$, $P(\myv a,\myv c)=-\cos\theta$, $P(\myv b,\myv c)=-\cos(\pi-\theta)$.
   

   Griffiths
   Chapter 4: 1: a only, 10, 11: a only, 13: a,b only, 16, 17

   Assignment #4 - due Friday, April 7

   Chapter 4: 22, 23, 24, 25, 27

   Problem 4.22, Construct $Y_l^l$ from ladder operators

   

   

   

   

   

   Problem 4.23, Applying the raising operator

   

   

   Problem 4.24, 2 masses connected by a rod

   

   

   Problem 4.27, Uncertainty relations for spin components

   In part c: use the the general property that $$\sigma^2_{\hat{Q}} =\langle(\hat{Q} -\langle\hat{Q}\rangle)^2\rangle = \langle\hat{Q}^2\rangle - \langle\hat{Q}\rangle ^2.$$
   
   

   

   

   

   Assignment #3 - due Wednesday, March 30

   Problem 4.17, Earth - Sun as quantized systemSequential measurements

   

   Wolfram Alpha calculation of Earth-sun Bohr radius

   

   

   

   

   Assignment #2 - due Monday, March 21

   Griffiths
   Chapter 3: 22, 27, 37, 39a
   Chapter 4: 2, 3, 5 (Just do $Y_3^2$)

    

   Assignment #1 (due Friday, March 11)

   Griffiths
   Chapter 3: 2, 6, 7, 13