# Comparing measurements of a 2D quantum harmonic oscillator between cartesian and rotated cartesian coordinates

I've come across an old quantum exam problem that's causing me a bit of confusion, and I'm hoping someone can offer some clarity:

There is a particle in a 2D harmonic oscillator potential such that it is described by the Hamiltonian (in unitless form):

$$H = -\frac{1}{2}(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}) + \frac{1}{2}(x^2+y^2)$$

Person A measures the system using x and y, while Person B uses a rotated coordinate system:

$$x' = x \: cos(\alpha) - y \: sin(\alpha)$$

$$y' = x \: sin(\alpha) + y \: cos(\alpha)$$

w/ $$\:\:0 < \alpha < \frac{\pi}{2}$$

The eigenstates for a 1D Harmonic Oscillator are:

$$\phi_0(x) = (\frac{1}{\pi})^{\frac{1}{4}}\: e^{-\frac{1}{2}x^2}$$

$$\phi_1(x) = (\frac{4}{\pi})^{\frac{1}{4}}\: x \: e^{-\frac{1}{2}x^2}$$

$$\phi_2(x) = (\frac{1}{4\pi})^{\frac{1}{4}}\: (2x^2-1) \: e^{-\frac{1}{2}x^2}$$

a) Is Person B's system still a two-dimensional harmonic oscillator?

To answer this, I inverted the expressions for x' and y' to get x and y in terms of x' and y' instead:

$$x = x' \: cos(\alpha) + y' \: sin(\alpha)$$

$$y = - x' \: sin(\alpha) + y' \: cos(\alpha)$$

And then replace the given Hamiltonian's x and y with these transformed versions:

$$H' = -\frac{1}{2} ([cos(\alpha)\frac{\partial}{\partial x'} + sin(\alpha)\frac{\partial}{\partial y'}]^2 +[-sin(\alpha)\frac{\partial}{\partial x'} + cos(\alpha)\frac{\partial}{\partial y'}]^2)$$

$$+ \frac{1}{2}([x' \: cos(\alpha) + y' \: sin(\alpha)]^2 +[- x' \: sin(\alpha) + y' \: cos(\alpha)]^2)$$

Doing some algebra, this yields:

$$H' = -\frac{1}{2}(\frac{\partial^2}{\partial x'^2}+\frac{\partial^2}{\partial y'^2}) + \frac{1}{2}(x'^2+y'^2)$$

(or we could just recognize that $$x^2 + y^2 = r^2$$, which is unchanged by rotation about the origin)

b) Person A puts the particle in state:

$$(n_x,n_y) = (1,0)$$ w/ $$E = 2\hbar\omega_0$$

Will Person B measure the same energy eigenvalue as Person A? Will Person B measure the same average energy as Person A?

First, I noted that $$E = 2\hbar\omega_0$$ is E = 2 in natural units. Using the given eigenstates, I computed:

$$\phi_{1,0}(x,y) = \phi_{1}(x)\phi_{0}(y) = (\frac{2}{\pi})^{\frac{1}{2}}\: x \: e^{-\frac{1}{2}(x^2+y^2)}$$

Then, to get this in terms of Person B's coordinates I again substituted x(x',y') and y(x',y'):

$$\phi'(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: [x' \: cos(\alpha) + y' \: sin(\alpha)] \: e^{-\frac{1}{2}(x'^2+y'^2)}$$

Then,

$$H'\phi' = 2 \cdot (\frac{2}{\pi})^{\frac{1}{2}}\: [x' \: cos(\alpha) + y' \: sin(\alpha)] \: e^{-\frac{1}{2}(x'^2+y'^2)}$$ (after a moderate amount of math)

Clearly, this indicates that:

$$H'\phi' = 2 \cdot \phi'$$

And so, $$E = 2\hbar\omega_0$$ is also an eigenvalue in the rotated system of Person B (this makes intuitive sense, as we wouldn't expect that measuring a system at a different angle would change measured values of energy). As it is an eigenvalue for Person B's system, it should be a guaranteed measurement in the prepared state (thus also being the average value of energy, and the same as Person A's).

c) Will Person B say that the system's particle is in an eigenstate?

I think this is where I'm starting to get a bit lost in the language. If my conclusion from b) is correct, then this must be 'yes', right? However, subsequent part d makes me question that this is so:

d) Determine the probability that Person B will say that the particle is in each of the following states:

1) $$(n_x',n_y') = (1,0)$$

2) $$(n_x',n_y') = (0,1)$$

3) $$(n_x',n_y') = (0,0)$$

If we've just measured the particle in an eigenstate of Person B's basis, then the probability of a subsequent measurement of the system resulting in a system in a different eigenstate must be zero, right?

What am I misunderstanding here? part d was worth nearly half of the problem's points, so it seems odd to result in such a trivial solution. Much appreciated!

It's probably not appropriate to have written:

$$\phi'(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: [x' \: cos(\alpha) + y' \: sin(\alpha)] \: e^{-\frac{1}{2}(x'^2+y'^2)}$$

Rather, since this is an eigenstate of the unprimed system but in terms of x' and y', it's probably better labeled just as $$\phi_{n_x = 1, n_y = 0}(x',y')$$

Since the primed system is also a harmonic oscillator, the degenerate eigenstates that produce E=2 would be:

$$\phi'_{1,0}(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: x' \: e^{-\frac{1}{2}(x'^2+y'^2)}$$

$$\phi'_{0,1}(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: y' \: e^{-\frac{1}{2}(x'^2+y'^2)}$$

So that the prepared state in terms of eigenstates of the primed basis is:

$$\phi_{1,0} = \cos\alpha \: \phi'_{1',0'} + \sin\alpha \: \phi'_{0',1'}$$

(The notation has become a bit convoluted, but hopefully you understand my meaning)

This result seems to make sense--- if $$\alpha = 0$$, we get back the original state, and if $$\alpha = \frac{\pi}{2}$$ we get the orthogonal (0',1') state instead.

Thank you again!

• Your answer to part (b) indeed tells you that Person B will measure the same energy as Person A. But Person B has two possible states that have that same energy. So it might be a superposition of the two! – octonion Mar 30 '19 at 21:59

The energy eigenvalue is known: it's 2, as you said, for B as well as for A. But energy eigenvalue $$E=2$$ has a 2D eigenspace, spanned by base vectors (1,0) and (0,1) both for B as for A. However these quantum numbers have different meaning for them: for A they refer to $$n_x$$, $$n_y$$ whereas for B they refer to $$n'_x$$, $$n'_y$$.
An observation of energy starting form state $$(n_x=1,n_y=0)$$ will certainly give an eigenvalue $$E=2$$ and the resulting state will be the projection of initial state in the subspace spanned by $$(n'_x=1,n'_y=0)$$ and $$(n'_x=0,n'_y=1)$$.
I leave for you to find that projection. (Hint: express $$n_x$$ as a linear combination of $$n'_x$$, $$n'_y$$.)
• Thank you! I've appended the new solution to my original question (as it was too long for a comment). My only follow-up, if you don't mind: do I remember correctly that, for degenerate eigenvalues, our resulting state is not either \phi'_{1',0'} or \phi'_{0',1'}, but rather exactly the linear combination of the two states --- $\phi_{1,0} = cos\alpha \: \phi'_{1',0'} + sin\alpha \: \phi'_{0',1'}$? – leo_africanus Apr 1 '19 at 15:22