Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Harmonic Oscillator position and momentum Hamiltonian operators

Posted by peeterjoot on December 18, 2010

[Click here for a PDF of this post with nicer formatting]

Motivation.

Hamiltonian problem from Chapter 9 of [1].

Problem 1.

Statement.

Assume x(t) and p(t) to be Heisenberg operators with x(0) = x_0 and p(0) = p_0. For a Hamiltonian corresponding to the harmonic oscillator show that

\begin{aligned}x(t) &= x_0 \cos \omega t + \frac{p_0}{m \omega} \sin \omega t \\ p(t) &= p_0 \cos \omega t - m \omega x_0 \sin \omega t.\end{aligned} \hspace{\stretch{1}}(3.1)

Solution.

Recall that the Hamiltonian operators were defined by factoring out the time evolution from a set of states

\begin{aligned}{\langle {\alpha(t) } \rvert} A {\lvert { \beta(t) } \rangle}={\langle {\alpha(0) } \rvert} e^{i H t/\hbar} A e^{-i H t/\hbar} {\lvert { \beta(0) } \rangle}.\end{aligned} \hspace{\stretch{1}}(3.3)

So one way to complete the task is to compute these exponential sandwiches. Recall from the appendix of chapter 10, that we have

\begin{aligned}e^A B e^{-A}= B + \left[{A},{B}\right]+ \frac{1}{{2!}} \left[{A},{\left[{A},{B}\right]}\right] + \cdots\end{aligned} \hspace{\stretch{1}}(3.4)

Perhaps there is also some smarter way to do this, but lets first try the obvious way.

Let’s summarize the variables we will work with

\begin{aligned}\alpha &= \sqrt{\frac{m \omega}{\hbar}} \\ X &= \frac{1}{{\alpha \sqrt{2}}} ( a + a^\dagger ) \\ P &= -i \hbar \frac{\alpha}{\sqrt{2}} ( a - a^\dagger ) \\ H &= \hbar \omega ( a^\dagger a + 1/2 ) \\ \left[{a},{a^\dagger}\right] &= 1 \end{aligned} \hspace{\stretch{1}}(3.5)

The operator in the exponential sandwich is

\begin{aligned}A = i H t/\hbar = i \omega t ( a^\dagger a + 1/2 )\end{aligned} \hspace{\stretch{1}}(3.10)

Note that the constant 1/2 factor will commute with all operators, which reduces the computation required

\begin{aligned}\antisymmetric{i H t/\hbar} {B } = (i\omega t) \left[{a^\dagger a},{B}\right]\end{aligned} \hspace{\stretch{1}}(3.11)

For B = X, or B = P, we’ll want some intermediate results

\begin{aligned}\left[{a^\dagger a},{a}\right]&=a^\dagger a a - a a^\dagger a \\  &=a^\dagger a a - (a^\dagger a + 1) a \\  &=-a,\end{aligned}

and

\begin{aligned}\left[{a^\dagger a},{a^\dagger}\right]&=a^\dagger a a^\dagger - a^\dagger a^\dagger a \\  &=a^\dagger a a^\dagger - a^\dagger (a a^\dagger -1) \\  &=a^\dagger\end{aligned}

Using these we can evaluate the commutators for the position and momentum operators. For position we have

\begin{aligned}\left[{i H t /\hbar },{X}\right] &= (i \omega t) \frac{1}{{\alpha \sqrt{2}}} \left[{a^\dagger a},{a+ a^\dagger}\right] \\ &= (i \omega t) \frac{1}{{\alpha \sqrt{2}}} (-a + a^\dagger ) \\ &= \frac{\omega t}{\alpha^2} \frac{-i \hbar \alpha}{ \sqrt{2}} (a - a^\dagger ).\end{aligned}

Since \alpha^2 \hbar = m \omega, we have

\begin{aligned}\left[{i H t /\hbar },{X}\right] = (\omega t) \frac{P}{m \omega }.\end{aligned} \hspace{\stretch{1}}(3.12)

For the momentum operator we have

\begin{aligned}\left[{i H t /\hbar },{P}\right] &= (i \omega t) \frac{-i \hbar \alpha}{ \sqrt{2}} \left[{a^\dagger a},{a- a^\dagger}\right] \\ &= (i \omega t) \frac{i \hbar \alpha}{ \sqrt{2}} (a + a^\dagger) \\ &= (\omega t) (\hbar \alpha^2) X\end{aligned}

So we have

\begin{aligned}\left[{i H t /\hbar },{P}\right] = (-\omega t) (m \omega ) X\end{aligned} \hspace{\stretch{1}}(3.13)

The expansion of the exponential series of nested commutators can now be written down by inspection and we get

\begin{aligned}X_H = X + (\omega t) \frac{P}{m \omega} - \frac{(\omega t)^2}{2!} X - \frac{(\omega t)^3}{3!} \frac{P}{m \omega} + \cdots\end{aligned} \hspace{\stretch{1}}(3.14)

\begin{aligned}P_H = P - (\omega t) (m \omega)X - \frac{(\omega t)^2}{2!} P + \frac{(\omega t)^3}{3!} (m \omega)X + \cdots\end{aligned} \hspace{\stretch{1}}(3.15)

Collection of terms gives us the desired answer

\begin{aligned}X_H = X \cos(\omega t) + \frac{P}{m \omega} \sin(\omega t)\end{aligned} \hspace{\stretch{1}}(3.16)

\begin{aligned}P_H = P \cos(\omega t) - (m \omega) X \sin(\omega t)\end{aligned} \hspace{\stretch{1}}(3.17)

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

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