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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!


Edit following answers:

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!

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    $\begingroup$ 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! $\endgroup$ – octonion Mar 30 '19 at 21:59
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There is only to elaborate @octonion's comment. Eigenvalues of energy are degenerate (ground state apart). Now A prepares state (1,0) and B measures energy. Which state will result from B's measurement?

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 $n'_x$, $n'_y$.)

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  • $\begingroup$ 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 --- $\phi_{1,0} = cos\alpha \: \phi'_{1',0'} + sin\alpha \: \phi'_{0',1'}$? $\endgroup$ – leo_africanus Apr 1 '19 at 15:22
  • $\begingroup$ OK. I've edited your last equation just to show you how to write trigonometric functions in MathJax (in TeX and LaTeX as well). $\endgroup$ – Elio Fabri Apr 1 '19 at 15:32
  • $\begingroup$ Noted. Thank you! $\endgroup$ – leo_africanus Apr 1 '19 at 23:48

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