Problem set 29

1. Let $\psi_{nlm}$ denote the eigenstates of a hydrogen atom in the usual notation. The state $\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]$ is an eigenstate of
1. $L^2$ but not of the Hamiltonian or $L_z$
2. the Hamiltonian, but not of $L^2$ or $L_z$
3. the Hamiltonian, $L^2$ and $L_z$
4. $L^2$ and $L_z$, but not of the Hamiltonian
$${\scriptstyle\Psi=\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]}$$ As $\psi_{nlm}$ are eigenfunctions of $H$, their linear combination will be eigenfunction of $H$. Now, $$\begin{align*} L_z\psi_{nlm}&=m\hbar\psi_{nlm} \text{ and } L^2\psi_{nlm}\\ &=l(l+1)\hbar^2\psi_{nlm} \end{align*}$$ $$\begin{align*} L_z\Psi&{\scriptstyle=L_z\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]}\\ &=\frac{1}{5}\left[-3\psi_{211}+\sqrt{5}\psi_{21-1}\right] \end{align*}$$ Hence, $\Psi$ is not eigenfunction of $L_z$. Similarly, one can show that $\Psi$ is not eigenfunction of $L^2$.
Hence, option (B) is correct.
2. The Hamiltonian for a spin-$1/2$ particle at rest is given by $H=E_0(\sigma_z+\alpha\sigma_x)$, where $\sigma_x$ and $\sigma_z$ are Pauli spin matrices and $E_0$ and $\alpha$ are constants. The eigenvalues of this Hamiltonian are
1. $\pm E_0\sqrt{1+\alpha^2}$
2. $\pm E_0\sqrt{1-\alpha^2}$
3. $E_0$ (doubly degenerate)
4. $E_0\left(1\pm\frac{1}{2}\alpha^2\right)$
$$\begin{align*} H&=E_0(\sigma_z+\alpha\sigma_x)\\ &=E_0\left\{\begin{pmatrix}1&0\\0&-1\end{pmatrix}+\alpha\begin{pmatrix}0&1\\1&0\end{pmatrix}\right\}\\ &=E_0\begin{pmatrix}1&\alpha\\\alpha&-1\end{pmatrix} \end{align*}$$ Eigenvalues of $H$ are obtained by solving $\det|H-\lambda I|=0$ $$E_0\begin{vmatrix}1-\lambda&\alpha\\\alpha&-1-\lambda\end{vmatrix}=0$$ $$\lambda=\pm E_0\sqrt{1+\alpha^2}$$ Hence, answer is (A)
3. For a system of independent non-interacting one-dimensional oscillators, the value of the free energy per oscillator, in the limit $T\rightarrow0$, is
1. $\frac{1}{2}\hbar\omega$
2. $\hbar\omega$
3. $\frac{3}{2}\hbar\omega$
4. $0$
(A) $\frac{1}{2}\hbar\omega$
4. If the reverse bias voltage of a silicon varactor is increased by a factor of 2, the corresponding transition capacitance
1. increases by a factor of $\sqrt{2}$
2. increases by a factor of $2$
3. decreases by a factor of $\sqrt{2}$
4. decreases by a factor of $2$
$$\begin{align*} C_T&\propto\sqrt{\frac{1}{V}}\Rightarrow\frac{C_T'}{C_T}\\ &=\sqrt{\frac{V}{V'}}\\ &=\sqrt{\frac{V}{2V}}\\ &\Rightarrow C_T'=\frac{1}{\sqrt{2}}C_T \end{align*}$$ Hence, answer is (C).
5. A cavity contains black body radiation in equilibrium at temperature $T$. The specific heat per unit volume of the photon gas in the cavity is of the form $C_v = \gamma T^3$,where $\gamma$ is a constant. The cavity is expanded to twice its original volume and then allowed to equilibrate at the same temperature $T$. The new internal energy per unit volume is
1. $4\gamma T^4$
2. $2\gamma T^4$
3. $\gamma T^4$
4. $\gamma T^4/4$
Change in internal energy $dU$ is given by $dU=C_vdT$ $$U=\int C_vdT=\int \gamma T^3dT=\frac{\gamma T^4}{4}$$ Hence, answer is (D)