Projecting an abstract state onto position/momentum/energy spaces

[v0id]

Homework Statement

Consider the quantum harmonic oscillator in the state $$| \psi (t) \rangle = \frac{1}{\sqrt{14}}\left( 3 | 0 \rangle \exp{\left( -\frac{1}{2}i \omega t\right)} + 2 | 1 \rangle \exp{\left( -\frac{3}{2}i \omega t\right)} + | 5 \rangle \exp{\left( -\frac{11}{2}i \omega t\right)} \right)$$. What is $$| \psi (t) \rangle$$ in terms of the $$\psi_n (x)$$, $$\psi_n (p)$$ and $$\psi_n (E)$$. Do not evaluate the specific basis vectors.

Homework Equations

$$\Phi (p) = \frac{1}{\sqrt{h}}\int_{-\infty}^{\infty} \psi (x) \exp{\left( \frac{-ipx}{\hbar} \right)} \; dx$$

The Attempt at a Solution

I think expansion along the energy space is unnecessary, since the original kets are themselves eigenkets of the Hamiltonian. As for position space, is the Fourier transform $$\Psi(x) = \frac{1}{\sqrt{h}} \int_{\infty}^{\infty} | \psi(t) \rangle \exp{\left( \frac{i \omega x}{\hbar}\right)} \; d\omega = \langle \phi | \psi(t) \rangle$$, where $$| \phi \rangle = \exp{\left( \frac{-i \omega x}{\hbar}\right)}$$?EDIT: Since only projection onto the position-space and momentum-space bases is necessary, would it be prudent to multiply the vector
$$\left[ \begin{array}{c} 3\\ 2\\ 0\\ 0\\ 0\\ 1 \end{array}\right]$$
by the matrix representations of the position and momentum operators respectively?

 

[Jhero]

Your approach is on the right track, but there are a few things to clarify.

First, the vector you have written is not the correct representation of the state | \psi (t) \rangle in the position basis. The correct representation would be a vector in the position basis, with elements given by the coefficients of the basis vectors in the state | \psi (t) \rangle. So, the correct vector would be:

\left[ \begin{array}{c} \frac{3}{\sqrt{14}} \exp{\left( -\frac{i\omega t}{2}\right)}\\
\frac{2}{\sqrt{14}} \exp{\left( -\frac{3i\omega t}{2}\right)}\\
0\\
0\\
0\\
\frac{1}{\sqrt{14}} \exp{\left( -\frac{11i\omega t}{2}\right)}
\end{array}\right]

Second, when you say "multiply the vector by the matrix representations of the position and momentum operators", I assume you mean taking the inner product between the vector and the matrix representations of the operators. This is correct, but you will need to do this for both position and momentum, and then combine the results to get the full representation of the state in terms of the position and momentum basis. So, your final result would be:

| \psi (t) \rangle = \left[ \begin{array}{c} \langle \psi_n (x) | \psi(t) \rangle\\
\langle \psi_n (p) | \psi(t) \rangle
\end{array}\right]

where \psi_n (x) and \psi_n (p) are the basis vectors in the position and momentum basis, respectively.