Phys 451 — Fall 2014 — Assignment 6

  1. A quantum particle confined to the \(x\)–\(y\) plane is subject to a central potential \(V(r) = V(\sqrt{x^2 + y^2})\). In polar coordinates (defined by \(x = r\cos\theta\) and \(y = r\sin\theta\)), the Hamiltonian

    \[\hat{H} = -\frac{\hbar^2 \nabla^2}{2m} + V(r)\]

    looks like

    \[\hat{H} = -\frac{\hbar^2}{2m}\biggl[ \frac{1}{r}\frac{\partial}{\partial r} \biggl( r \frac{\partial}{\partial r}\biggr) + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\biggr] + V(r).\]

    Express the transformed wave function \(\psi(x,y) \to \psi(r,\theta)\) as the product \(\psi(r,\theta) = R(r)\Theta(\theta)\). Show that doing so leads to two independent eigen-equations for \(R(r)\) and \(\Theta(\theta)\).

  2. The angular momentum operator (a scalar rather than a vector in 2D) is

    \[\hat{L} = \mathbf{e}_z \cdot (\mathbf{r} \times \mathbf{p}) = - i \hbar \mathbf{e}_z \cdot (\mathbf{r} \times \nabla).\]

    Show that both of the following representations are correct:

    \[\hat{L} = \frac{\hbar}{i}\biggl( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \biggr) = \frac{\hbar}{i} \frac{\partial}{\partial \theta}.\]

  3. Prove that \([\hat{H},\hat{L}] = 0\).

  4. By virtue of the result in question 3, we know that the wavefunction can be a simultaneous eigenfunction of the Hamiltonian and the angular momentum. Solve for the states of definite angular momentum,

    \[\hat{L}\Theta_m(\theta) = \hbar m \Theta_m(\theta).\]

    What is the functional form of \(\Theta_m(\theta)\), and what values can \(m\) take?

  5. Consider the specific example of a quantum harmonic oscillator with

    \[V(r) = \frac{1}{2}m\omega^2r^2 = \frac{1}{2}m\omega^2(x^2+y^2).\]

    Argue that the Hamiltonian can be expressed as $$\hat{H} = \hbar\omega(a^\dagger a

    • b^\dagger b + 1)\(, where\)a\(and\)b\(are lowering operators in each of the\)x\(and\)y$$ directions.

  6. Use the usual definitions,

    \[\begin{align} \hat{x} &= \sqrt{\frac{\hbar}{2m\omega}} (a^\dagger+a)\\ \hat{y} &= \sqrt{\frac{\hbar}{2m\omega}} (b^\dagger+b)\\ \hat{p}_x &= i\sqrt{\frac{m\hbar\omega}{2}} (a^\dagger-a)\\ \hat{p}_y &= i\sqrt{\frac{m\hbar\omega}{2}} (b^\dagger-b), \end{align}\]

    to show that

    \[\hat{L} = i\hbar(a b^\dagger - a^\dagger b).\]

  7. Again, prove that \([\hat{H},\hat{L}] = 0\), but this time do so by manipulating the raising and lowering operators.

  8. Consider a basis \(\{ \lvert 0,0 \rangle, \lvert 1,0 \rangle, \lvert 0,1 \rangle, \lvert 2,0 \rangle, \lvert 1,1 \rangle, \lvert 0,2 \rangle, \ldots \}\) consisting of states \(\lvert n_x,n_y \rangle\) of definite number:

    \[\begin{align} a^\dagger a \lvert n_x,n_y \rangle = n_x \lvert n_x,n_y \rangle,\\ b^\dagger b \lvert n_x,n_y \rangle = n_y \lvert n_x,n_y \rangle. \end{align}\]

    Expressed in this basis, the Hamiltonian is diagonal.

    \[H = \begin{pmatrix} \hbar \omega \\ & 2\hbar \omega & 0\\ & 0 & 2\hbar \omega \\ &&& 3\hbar \omega & 0 & 0 \\ &&& 0 & 3\hbar \omega & 0 \\ &&& 0 & 0 & 3\hbar \omega \\ &&&&&& \ddots \\ \end{pmatrix}.\]

    The angular momentum, however, is not.

    \[L = \hbar \begin{pmatrix} 0 \\ & 0 & 1\\ & 1 & 0 \\ &&& ? & ? & ? \\ &&& ? & ? & ? \\ &&& ? & ? & ? \\ &&&&&& \ddots \\ \end{pmatrix}.\]

    Compute matrix elements to fill in the nine missing entries.

  9. Construct states with energy \(\hbar\omega\), \(2\hbar\omega\), and \(3\hbar\omega\) that are also states of definite angular momentum. Proceed by diagonalizing the \(1\times 1\), \(2 \times 2\), and \(3 \times 3\) blocks appearing in the matrices in question 8.