hope you're all doing well.

There might be questions related to this topic but I couldn't stand other responses mainly because I just began learning QM. There's this problem on QM1 (Quantum Mechanics at Undergraduate level, Griffiths QM-ish) in which I can't interpretate the result.

I'm asked to find the probability density distribution for a particle trapped in an infinite square well (so it's a free unbounded particle between 0 and d with Dirichlet boundary conditions).

Due to Plancherel's theorem (p.60 Griffiths) I know I can represent the wave in terms of the wave number using Fourier transform:

$$\chi(k) = \frac{1}{\sqrt{2\pi}}\int \Psi(x,0) e^{-ikd} \mathrm{d}x$$

At this point I also know that the time-independent solution to the square well coincides with $\Psi_0 (x)$, which emerges from the ODE $\phi_n '' + k\phi_n =0$. The integral becomes

$$\chi(k) = \frac{1}{\sqrt{2\pi}}\int \phi(x) e^{-ikd} \mathrm{d}x = \frac{1}{\sqrt{2\pi}}\int \sqrt{\frac{2}{d}} \sin\left(\frac{n \pi x}{d}\right) e^{-ikd} \mathrm{d}x $$

Integrated in the $x$ domain, $[0,d]$. I calculated the integrals by parts (work which I can copy later if needed) and got to:

$$\chi(k) = \frac{\sqrt{2} e^{-i k d} (-e^{i k d} n \pi + n \pi (-1)^n)}{\sqrt{\frac{1}{d}}(d^2 k ^2 - n^2 \pi^2)}$$

Next step was taking $\|\chi(k)\|^2=\chi^* \chi$ and the result was

$$p(k) \equiv \|\chi(k)\|^2 = \frac{4dn^2\pi^2}{(d^2k^2 - n^2\pi^2)^2}(1-(-1)^n \cos(d k))$$

So my first question is:

  • The $\chi(k)$ value I calculated depends on n. The correct way to do this should be the sum of n from 1 to $\infty$ ? How should I compute $p(k)$ then?
  • What is representing this $p(k)$ and how is this related to the Fourier transform?
  • Should $p(k)$ be normalized?
  • If I plot this p(k) for a fixed d=1, with respect to k from 0.5 to 100, using Wolfram Mathematica
P[k_, n_, d_] = (
  4 d n^2 \[Pi]^2)/(d^2 k^2 - n^2 \[Pi]^2)^2 (1 - (-1)^n Cos[d k])
Animate[Plot[P[k, n, 1], {k, 0.5, 100}, PlotRange -> All, 
  PlotTheme -> "Detailed", PlotStyle -> Red], {n, 1, 50, 1}]

I get

p(k) at an arbitrary n1 p(k) at an arbitrary n2>n1 p(k) at an arbitrary n3>n2 in which the peak of p(k) is at k=d p(k) what happens now? and now p(k) tends to a sine wave

  • Why is this happening? Why $p(k)$ behaves this way?
  • And if I sum the first 7 states this is what I get. Why?

sum of the first p_n(k) from n= 1 to 7

I apologize if I'm being a little vague with my questions but I seriously don't know how to interpret this.

  • $\begingroup$ It's unclear to me what you're trying to do here. You need to know $\Psi(x)$ before computing its Fourier transform $\chi(k)$. In your question, you've specifically chosen $\Psi(x)$ to be the $n^{th}$ energy eigenstate (which is why $\chi(k)$ depends on $n$) but it sounds like this was not your intention. $\endgroup$
    – J. Murray
    Sep 3 '20 at 0:59
  • $\begingroup$ @J.Murray I've chosen the this function because I'm asked to determine the probability density in the momentum space for the fundamental state and the first four next states in the context of a particle trapped in a 1D square well (w/ infinite potential), so $Psi(x)$ is solution of this particular problem. $\endgroup$
    – nuwe
    Sep 4 '20 at 16:37
  • $\begingroup$ I'm not sure why you're trying to add them together for different $n$, then. You have five different answers ($n=1,2,3,4,5$) corresponding to the five states you are supposed to consider. $\endgroup$
    – J. Murray
    Sep 5 '20 at 0:13
  • why do you sum over $n$ ?. you had chosen $\Psi(x,0)$ to be the $n$th eigenstates.

  • Parseval's relation (or perhaps it has another name) $$ \int_\mathbb{R} f^*(x)f(x)dx = \int_\mathbb{R} \tilde{f} ^*(k) \tilde{f}(k)dk $$ tells that if $f(x)$ is normalized then its Fourier tranform $\tilde{f} ^*(k)$ is as well.

  • why do you use $||\ \ ||^2$ ?, this is norm sign, it means here $$ ||\Psi||^2= \langle \Psi| \Psi \rangle = \int_\mathbb{R} \Psi^*(x,t) \Psi(x,t)\ dx $$

  • $\begingroup$ hi, I sum over $n$ to see what would be an estimated behavior of the superposition of states. Anyways I know this is not computed accurately and I should weight each function by its probability. Please I ask for correction if I'm wrong. Concerning on the norm, I'm actually not sure if I should compute as the product between conjugates or the integral of this. Could you clear this out? I don't understand quite much about why I should integrate it and why this shouldn't result on being just 1 (because of the normalization and Parseval's relation). $\endgroup$
    – nuwe
    Sep 4 '20 at 16:40
  • $\begingroup$ Wave function is a vector on Hilbert space, which is, technically speaking, a complete inner product space. Not every vector in Hilbert space is state vector (wave function) but any such vector can be normalized to get possible state vector. The inner product in this Hilbert space is what induces the norm. The Hilbert space in which wave functions live is, roughly speaking, the set of all squareintegrable functions. The inner product on this space is theintegral like the one in the parseval's relation. The norm on this space is induced by the inner product, as in the answer $\endgroup$
    – Physor
    Sep 4 '20 at 18:17
  • $\begingroup$ I suppose you know about the wave function and the probabilty distribution t represents if you read from Griffiths. That requires the norm of every wave function to 1. $\endgroup$
    – Physor
    Sep 4 '20 at 18:21
  • $\begingroup$ about the first thing in your first comment. I don't really understand what you expect from summing over $n$. could you explain why do you do that ? $\endgroup$
    – Physor
    Sep 4 '20 at 18:22

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