Expectation value of $ Y \otimes I \otimes I $ for a charged particle in a magnetic field

Question

Typically when solving a Hamiltonian of ye olde form $ \frac{1}{2m} (\bar P - \frac{q}{c} \bar A)^2 $, you do separation of variables.

For simplicity say that $ A = - B Y \hat x $.

You can rewrite it as $\frac{1}{2m} ( (\hbar k_x + \frac{q}{c} B y )^2 + P_y^2 + \hbar^2 k_z^2 ) $.

Now, we're putting everything in the position basis by applying $ \langle x | \otimes \langle y | \otimes \langle z |$ on the left side, and then theoretically some sort of $ |\psi\rangle $ in 3 dimensions on the right side, and then computing the results.

However, then I'm slightly confused about what happens formally with the equation for the y, $ (\hbar k_x + \frac{qB}{c} y)^2 + P_y^2 $. You can solve this as a harmonic oscillator and you'll get solutions for $ \psi(y)$.

Now say, for instance, say you wanted to compute $ \langle Y \otimes I \otimes I\rangle $. In theory, you'd miss it altogether if you apply $ | \psi(x) \rangle \otimes | \psi(y) \rangle \otimes |\psi(z) \rangle $ since Y will only operate on $ \psi(x)$ which is just a plane wave. Frankly I'm not even sure how this computation would work/if it'll just pass through or something. You'd have to apply something like $ | \psi(y) \rangle \otimes | \psi(x) \rangle... $ and so on.

But this shouldn't occur because you should be able to somehow measure the average "Y" value to be $ \frac{\hbar k_x c}{qB} $. This also leads to bad outcomes because if you decide to compute $ \langle dX/dt \rangle $, you'd compute the commutator of $ X \otimes I \otimes I $ with the Hamiltonian; it'll commute with anything which is the identity in the X coordinate; $ D_t \langle X \rangle $ will end up being proportional to the expectation value of Y, and I know this quantity shouldn't end up being trivial or something.

Can someone point out the misstep here in how I propose to find the expectation value for Y?

Answer 1

I think your main confusion is that you are treating the Hilbert space as having a tensor product structure, with the different tensor factors corresponding to the different coordinates in position space. However, the Hilbert space does not decompose like that, i.e. $\mathcal{H} \neq \mathcal{H}_x \otimes \mathcal{H}_y \otimes \mathcal{H}_y$. This observation applies to all of the operators, states, wavefunctions, etc. (Note that while the wavefunction $\psi(x,y,z)$ may be separable as $\psi(x,y,z) = \phi(x)\chi(y)\zeta(z)$, this is not the same as saying $\langle x, y, z|\psi\rangle = \left(\langle x| \otimes \langle y| \otimes \langle z|\right)|\psi\rangle$)

In particular, when you write the operator $\vec{A} = -By\,\hat{x}$, the $\hat{x}$ is just telling us about how $\vec{A}$ looks as an ordinary Euclidean vector, not anything to do with tensor factors.

If you want to find the expectation value for the operator $y$, you can do it into the position basis like so: \begin{align} \langle y \rangle &= \langle\psi|y|\psi\rangle\\ &= \int d^3\vec{r}\, y\,\psi^{\ast}(x,y,z)\psi(x,y,z)\\ &= \int d^3\vec{r}\, y\,|\psi(x,y,z)|^2 \end{align} Your wavefunction will like look $\psi(x,y,z) = A e^{-ik_x x - ik_z z}\chi(y)$ so you will end up with \begin{align} \langle y \rangle \propto \int dy\, y\,|\chi(y)|^2, \end{align} which you should be able to do with standard integration techniques.