Math formula representation example

An example containing mathematica formula

PHYSICS DEPARTMENT

PHY 1520F QUANTUM MECHANICS

Assignment No. 2 - Solution

Problem 1

The Hamiltonian of a free particle is \[ H = {1\over 2m} \sum^3_{i-1} p^2_i.\label{2solp1e1}\]

The Heisenberg equations of motion for \(p_i\) and \(x_j\) are \[{dp_j\over dt} = {1\over i\hbar} [p_j, H] = 0,\label{2solp1e2}\] \[ {dx_j\over dt} = {1\over i\hbar}[x_j, H] = {1\over 2i\hbar m}\sum^3_{i=1}[x_j, p^2_j] = {p_i\over m},\label{2solp1e3}\]

From Eq. {2solp1e2} we have \(p_j(t) - p_j(0)\).

Hence the solution of Eq.{2solp1e2} is \[ x_j(t) = x_j(0) + {p_j(0)\over m}t.\label{2solp1e4}\]

Since \([x_i, p_j] = i\hbar\delta_{ij}\) we obtain \[ [x_j(t), x_i(0)] = -{i\hbar t\over m}\delta_{ij}.\label{2solp1e5}\] \setcounter{equation}{0}

Problem 2

The Hamiltonian of a one-dimensional harmonic oscillator is \[ H = {p^2\over 2m} + {m\omega^2x^2\over 2} = \hbar\omega(a^\dagger a + 1),\label{2solp2e1}\] where \[ x = \left({\hbar\over 2m\omega}\right)^{1/2} (a + a^\dagger),\label{2solp2e2}\] \[ p = i \left({m\hbar\omega\over 2}\right)^{1/2} (-a+a^\dagger).\label{2solp2e3}\]

The Heisenberg equation of motion for operator \(a\) can be found as \[{d\over dt} a = {1\over i\hbar}[a, H] = -i\omega a.\label{2solp2e4}\]

The solution of Eq.{2solp2e4} is \[ a(t) = a(0) e^{-i\omega t},\label{2solp2e5}\] and likewise \[ a^\dagger (t) = a^\dagger (0) e^{i\omega t}.\label{2solp2e6}\]

From Eqs.{2solp2e2})-{2solp2e3}) and {2solp2e5}-{2solp2e6} one has \[ x(t) = \left({\hbar\over 2m\omega}\right)^{1/2}\left[a(0)e^{-i\omega t}+a^\dagger(0)e^{i\omega t}\right] = x(0)\cos(\omega t) + {p(0)\over m\omega}\sin(\omega t).\label{2solp2e7}\]

With the help of Eq. (\ref{2solp2e7}) one can find the expectation value \(\langle x(t) \rangle\) over the initial state \(e^{-ipb/\hbar}\vert 0 \rangle\) as \[\langle x(t) \rangle = \langle 0\vert e^{ipb/\hbar} x(0) e^{-ipb/\hbar}\vert 0 \rangle \cos (\omega t) + {\sin(\omega t)\over m\omega} \langle 0\vert p(0)\vert 0 \rangle.\label{2solp2e8}\]

Since \(\langle 0\vert p(0)\vert 0 \rangle = 0,\) Eq. (\ref{2solp2e8}) becomes \[ \langle x(t) \rangle = \langle 0\vert e^{ipb/\hbar}i\hbar{\partial\over \partial p} e^{-ipb/\hbar}\vert 0 \rangle \cos (\omega t),\label{2solp2e9}\] which yields \[ \langle x(t) \rangle = b\cos (\omega t).\label{2solp2e10}\]