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Consider a free particle in a uniform magnetic field, Hamiltonian $H = ( \mathbf p - e \mathbf A)^2/2m$ and magnetic field in the $z$ direction.

  • Solving the Schrödinger equation with $\mathbf A = (-B_y,0,0)$, I get solutions $$\Psi_a(x,y) = e^{ik_xx} e^{-(y+y_k)/(2l^2)}$$ for the ground state, with $y_k = l^2 k_y $ and $l^2 = (\hbar c )/ (e B) $.

  • On the other hand, If I solve the equation with $\mathbf A = (0,B_x,0)$, I get as solutions $$\Psi_b(x,y) = e^{ik_yy} e^{-(x-x_k)/(2l^2)},$$ with $x_k = l^2 x_y $.

I am now asked to show that I can take linear combinations of states localized in $y$ ($\Psi_a$) and obtain states localized in the $x$ direction ($\Psi_b$). I am confused about how to do this since I don't see how a sum over exponential functions localized in one direction can add up to an exponential function(s) localized in other direction. It has to be possible since both cases represent the same physical state so the solutions must be equivalent. Do you know how can this be done?

Emilio Pisanty
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QPhysJP
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1 Answers1

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This is the essence of Fourier transforms: when you write $$ e^{-x^2/2\sigma^2} = \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12 \sigma^2k^2}e^{ikx}\mathrm dk, $$ you're expressing the localized function $e^{-x^2/2\sigma^2}$ as an infinite linear combination of the delocalized functions $e^{ikx}$.

A word of warning, though: your conclusion in

both cases represent the same physical state so the solutions must be equivalent

isn't quite right. The fact that both cases represent the same physical state tells you that

  1. any solution of case (a) must also be a solution of case (b), and in this case also that
  2. for any solution of (a), there exists a solution of (b) of the same form, and more generically that
  3. any solution of (a) can be expressed as a linear combination of the natural solutions of (b)

but it doesn't tell you that any natural solution of (a) is a natural solution of (b).

(As a simple example of that form, consider the translationally- and rotationally-invariant solutions for a free particle in 2D. They're both equivalent systems, they both admit nice solutions in the form of plane waves and Bessel waves, and both of their solutions are superpositions of the other ones, but plane waves aren't Bessel waves or vice versa.)

Emilio Pisanty
  • 137,480