Question

Consider the wave function 𝜓(𝑥) = 𝐴𝑐𝑜𝑠 𝑘𝑥 where A, k are constants. Is this the eigenstate of the operators 𝐻̂, 𝑝̂, 𝑝̂ 2 of a particle moving in a potential-free region. If so find the eigenvalues.

Answer 1

To verify it, we will simply apply these operators (we know their $x$ representation) :

• $\hat{p}\psi=-i\hbar \frac{d}{dx}\psi = i\hbar Ak \sin(kx)$ . We clearly see that $\psi, \hat{p}\psi$ are not collinear (as sine and cosine are not proportional to each other), so $\psi$ is not an eigenstate of $\hat{p}$
• $\hat{p}^2\psi = \hat{p}(\hat{p}\psi) = -i\hbar\frac{d}{dx}(i\hbar Ak\sin(kx)) = A \hbar^2k^2\cos(kx) =\hbar^2k^2\psi$ . So $\psi$ is an eigenstate of $\hat{p}^2$ with an eigenvalue $\hbar^2k^2$ .
• $\hat{H} \psi = (\frac{\hat{p}^2}{2m}+\hat{V})\psi =\frac{1}{2m} \hat{p}^2\psi = \frac{\hbar^2k^2}{2m}\psi$ . Again $\psi$ is an eigenstate of $\hat{H}$ with an eigenvalue $\frac{\hbar^2k^2}{2m}$ .