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Problem: In the context of quantum harmonic oscillator the eigenfunctions are given by: $ u_n(x) = (N_n/\sqrt{b}) H_n(x/b) \exp\left[-x^2/(2b^2)\right] $, where $N_n$ is the normalization factor: $ N_n = \left(\frac{1}{\sqrt{\pi} 2^n n!}\right)^{1/2} $ and $b = \left(\frac{\hbar^2}{m k}\right)^{1/4}$, where $m$ is the mass and $k$ is the spring constant.

My book asks me to prove that the following equation holds, where $\varphi_n(k)$ are the eigenfunctions in the momentum space:

$ \varphi_n(k) = \sqrt{b} N_n H_n(b k) \exp(-b^2 k^2 / 2), k=p/\hbar $

Attempted solution:

norm[n_] := (1/(Sqrt[π] 2^n n!))^(1/2)

u[n_, x_] := 1/Sqrt[b] norm[n] HermiteH[n, x/b] Exp[-(x^2/(2 b^2))]

φ1[n_, k_] :=
  {n ∈ Integers, n > 0, x ∈ Reals, k ∈ Reals, b > 0},
  1/Sqrt[2 π] Integrate[u[n,x]Exp[-I k x], {x, -∞, +∞}]]

φ2[n_, k_] :=
  {n ∈ Integers, n > 0, b > 0},
  Sqrt[b] norm[n] HermiteH[n, b k] Exp[-((b^2 k^2)/2)]]

    {j, Simplify[φ1[j, k] == φ2[j, k]]},
    {j, 1, 8}] // TableForm

And here is the output:

enter image description here

My question is: Why does the relation hold only for $4k, k \in Z$ ? If I print $φ1[n,k]$ and $φ2[n,k]$ I see that in $φ1$ there are $(-i)^j$ terms, which is why the equation holds only for multiples of 4.

Does anyone have any idea at all on where the error is ? (In my book? In how I do the Fourier transform? Something else I'm missing?)

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Hi Zet, I think this question looks more likely belong to the Physics site. – Silvia Dec 10 '13 at 4:05
up vote 6 down vote accepted

Your code is ok. What is missing is the phase factor in the definition of momentum-space wavefunction $\varphi_2(n,k)$ that you use. The proper definition is:

$\varphi_2(n,k) = (-i)^n \sqrt{b} N_n H_n(b k) \exp(-b^2 k^2 / 2) ,$

as can be verified for example in these lecture notes.

Also, I've found that a slight change in your testing function is helpful in making the comparisons:

Table[{j, Expand[φ1[j, k]] == Expand[φ2[j, k]]}, {j, 1, 8}] // TableForm

Note also that for a quantum harmonic oscillator, the possible values of the quantum number $n$ are $n = 0,1,2,3,\ldots$, so you probably want also to check the equality for $n=0$.

share|improve this answer
Thanks @au700!! – Zet Dec 10 '13 at 17:01

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