Sample Test Problems

1. Derive the commutators $[L^2,L_+]$ and $[L_z,L_+]$. Now show that $L_+Y_{\ell m}=CY_{\ell (m+1)}$, that is, $L_+$ raises the $L_z$ eigenvalue but does not change the $L^2$ eigenvalue.
   Answer
   
   
   $$L_+=L_x+iL_y$$
   
   
   Since $L^2$ commutes with both $L_x$ and $L_y$,
   
   $$[L^2,L_+]=0 .$$
   
   
   
   
   $$[L_z,L_+]=[L_z,L_x+iL_y]=[L_z,L_x]+i[L_z,L_y]=i\hbar(L_y-iL_x)=\hbar(L_x+iL_y)=\hbar L_+$$
   
   
   We have the commutators. Now we apply them to a $Y_{\ell m}$.
   
   $$[L^2,L_+]Y_{\ell m}=L^2L_+Y_{\ell m}-L_+L^2Y_{\ell m}=0$$
   
   
   
   
   $$L^2(L_+Y_{\ell m})=\ell(\ell+1)\hbar^2(L_+Y_{\ell m})$$
   
   
   So, $L_+Y_{\ell m}$ is also an eigenfunction of $L^2$ with the same eigenvalue. $L_+$ does not change $\ell$.
   
   $$[L_z,L_+]Y_{\ell m}=L_zL_+Y_{\ell m}-L_+L_zY_{\ell m}=\hbar L_+Y_{\ell m}$$
   
   
   
   
   $$L_z(L_+Y_{\ell m})-m\hbar (L_+Y_{\ell m})=\hbar (L_+Y_{\ell m})$$
   
   
   
   
   $$L_z(L_+Y_{\ell m})=(m+1)\hbar (L_+Y_{\ell m})$$
   
   
   So, $L_+$ raises the eigenvalue of $L_z$.
2. Write the (normalized) state which is an eigenstate of $L^2$ with eigenvalue $\ell(\ell +1)\hbar^2=2\hbar^2$ and also an eigenstate of $L_x$ with eigenvalue $0\hbar$ in terms of the usual $Y_{\ell m}$.
   Answer
   An eigenvalue of $\ell(\ell +1)\hbar^2=2\hbar^2$ implies $\ell =1$. We will need a linear combination of the $Y_{1m}$ to get the eigenstate of $L_x={L_++L_-\over 2}$.
   
   $${L_++L_-\over 2}(AY_{11}+BY_{10}+CY_{1-1})=0$$
   
   
   
   
   $${\hbar\over\sqrt{2}}(AY_{10}+BY_{11}+BY_{1-1}+CY_{10})=0$$
   
   
   
   
   $$(BY_{11}+(A+C)Y_{10}+BY_{1-1})=0$$
   
   
   Since this is true for all $\theta$ and $\phi$, each term must be zero.
   
   $$B=0$$
   
   
   
   
   $$A=-C$$
   
   
   The state is
   
   $${1\over\sqrt{2}}(Y_{11}-Y_{1-1})$$
   
   
   The trivial solution that $A=B=C=0$ is just a zero state, not normalizable to 1.
3. Write the (normalized) state which is an eigenstate of $L^2$ with eigenvalue $\ell(\ell +1)\hbar^2=2\hbar^2$ and also an eigenstate of $L_y$ with eigenvalue $1\hbar$ in terms of the usual $Y_{\ell m}$.
4. Calculate the commutators $[p_z,L_x]$ and $[L_x^2,L_z]$.
5. Derive the relation $L_+Y_{\ell m}=\hbar\sqrt{\ell(\ell+1)-m(m+1)}Y_{\ell (m+1)}$.
6. A particle is in a $l=1$ state and is known to have angular momentum in the $x$ direction equal to $+\hbar$. That is $L_x\psi=\hbar\psi$. Since we know $l=1$, $\psi$ must have the form $\psi=R(r)(aY_{1,1}+bY_{1,0}+cY_{1,-1})$. Find the coefficients $a,b,$ and $c$ for $\psi$ normalized.
7. Calculate the following commutators: $[x,L_z]$, $[L_+,L^2]$, $[{1\over 2}m\omega^2r^2,p_x]$.
8. Prove that, if the Hamiltonian is symmetric under rotations, then $[H,L_z]=0$.
9. In 3 dimensions, a particle is in the state:
   
   $$\psi({\bf r}) = C(iY_{33}-2Y_{30}+3Y_{31}) R(r)$$
   
   
   where $R(r)$ is some arbitrary radial wave function normalized such that
   
   $$\int_{0}^{\infty}R^* (r) R(r) r^2 dr = 1 .$$
   
   

   a)

   Find the value of C that will normalize this wave function.

   b)

   If a measurement of $L_z$ is made, what are the possible measured values and what are probabilities for each.

   c)

   Find the expected value of $<L_x>$ in the above state.

10. Two (different) atoms of masses $M_1$ and $M_2$ are bound together into the ground state of a diatomic molecule. The binding is such that radial excitations can be neglected at low energy and that the atoms can be assumed to be a constant distance $a=3\AA$ apart. (We will ignore the small spread around $r=a$.)

    a)

    What is the energy spectrum due to rotations of the molecule?

    b)

    Assuming that $R(r)$ is given, write down the energy eigenfunctions for the ground state and the first excited state.

    c)

    Assuming that both masses are about 1000 MeV, how does the excitation energy of the first excited state compare to thermal energies at $300^{\circ}$K.