Position and Momentum Operator Matrices

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 13K views
Messages
665
Reaction score
68

Homework Statement


Find the energy eigenvalues and eigenfunctions for the one-dimensional infinite square well. Calculate the matrices for the position and momentum operators, Q and P, using these eigenfunctions as a basis.

Homework Equations


The energy eigenvalues are

[tex]E_n = \frac{\pi^2 \hbar^2}{2Ma^2}n^2[/tex]

and the eigenfunctions are,

[tex]\psi_n(x) = \sqrt{\frac{2}{a}} sin \left ( \frac{n \pi}{a} x \right )[/tex]

The Attempt at a Solution



Finding the energy eigenvalues and eigenfunctions is straightforward. What is really confusing me is how to use the eigenfunctions as a basis to find the matrix representations of these operators. This is problem 4.2 in Ballentine's Quantum Mechanics, and I can't find any useful information in the chapter for doing this sort of problem. Here is the best attempt I have:

The position operator Q satisfies,

[tex]\hat Q \psi_n(x) = x_n \psi_n(x)[/tex]

If we represent [itex]\psi[/itex] as a column matrix, we can write:

[tex]\psi_n(x) = \left[ \begin{array}{c} \psi_1(x) \\ \psi_2(x) \\ \vdots \\ \psi_n(x) \end{array} \right][/tex]

Using this representation,

[tex]\hat Q \left[ \begin{array}{c} \psi_1(x) \\ \psi_2(x) \\ \vdots \\ \psi_n(x) \end{array} \right] = \left[ \begin{array}{c} x_1 \psi_1(x) \\ x_2 \psi_2(x) \\ \vdots \\ x_n \psi_n(x) \end{array} \right]= \left[ \begin{array}{cccc} x_1 & 0 & \hdots & 0\\ 0 & x_2 & 0 \hdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & x_n \end{array} \right] \left[ \begin{array}{c} \psi_1(x) \\ \psi_2(x) \\ \vdots \\ \psi_n(x) \end{array} \right][/tex]

So Q would be this matrix with the x's along the diagonal:

[tex]\hat Q = \left[ \begin{array}{cccc} x_1 & 0 & \hdots & 0\\ 0 & x_2 & 0 \hdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & x_n \end{array} \right][/tex]

I feel like the answer is supposed to be more specific than this
 
Last edited:
Physics news on Phys.org
kreil said:
The position operator Q satisfies,

[tex]\hat Q \psi_n(x) = x_n \psi_n(x)[/tex]

If you are talking about a one-dimensional harmonic oscillator, what the heck is [itex]x_n[/itex] supposed to mean? In one dimension, [itex]\hat{Q}\psi_n(x)=x\psi_n(x)[/itex].

The elements of a matrix [itex]A[/itex] in a basis [itex]|v_i\rangle[/itex] are given by

[tex]A_{ij}= \langle v_i|A|v_j\rangle[/tex]

Apply that to find the matrix elements of the position operator in the [itex]\psi_n\rangle[/itex] eigenbasis (Hint: [itex]\psi_n(x)=\langle x|\psi_n\rangle[/itex] and [tex]\int |x\rangle\langle x|dx = \hat{1}[/tex] ).