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!
