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Consider a charged quantum particle confined to the $xy$ plane, subject to a magnetic field $\mathbf{B}=B\hat{z}$.
The Hamiltonian is:

$$ H = \frac{1}{2m} \left( \mathbf{p} - \frac{e \mathbf{A}}{c}\right)^2~, $$

where $\mathbf{A}$ is the vector potential, for which we have a freedom of gauge in choosing. One possible choice is $\mathbf{A} = B x \hat{y}$, which leads to the Hamiltonian:

$$ H = \frac{1}{2m} \left[ p_x^2 + \left(p_y- m \omega_c x\right)^2\right],$$

where $\omega_c = eB/mc$ is the cyclotron frequency. Following the usual derivation of Landau quantization, we get the wavefunctions:

$$ \psi(x,y) = f_n ( x- k_y / m \omega_c ) e^{i k_y y},$$

where $f_n$ are the eigenfunctions of the simple harmonic oscillator ($n=0,1,2...$).

However, we can also choose $\mathbf{A} = -B y \hat{x}$, which following the same lines would lead to the wavefunctions:

$$ \psi(x,y) = f_n ( y+ k_x / m \omega_c ) e^{i k_x x}.$$

The two outcomes seems different by more then just a global phase, yet all we did was use our freedom in picking a gauge.
How does this work out?

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up vote 1 down vote accepted

yet all we did was use our freedom in picking a gauge.

No, you did another thing: choosing the complete set of commuting observables. In the first case you choose $\left\{H,p_y\right\}$ and the second $\left\{H,p_x\right\}$. The difference in operators ($p_x$ vs $p_y$) leads to difference in eigenvectors.

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And the two sets of eigenvectors, when taken for the same energy, are connected by a linear transformation, like @F.Solis said in his answer. The two answers complete each other and together make up the full answer, I'm accepting this one just because it was first... – Joe Feb 1 '13 at 19:28

In addition to the comments of Joe, note that both sets of eigenvectors lead to the same energy spectrum. This is consistent with the fact that both of your sets properly characterize the same physical system. The two sets are connected by a complicated linear transformation.

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