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I'm learning quantum physics by my own, and I can't completely understand the information given by a wave function.

My problem came with the solution of the one-dimensional box problem. So, supposing that we all know the conditions of this famous problem, we can calculate the wave function using the time-independent Schrödinger's equation:

$$ \hat{H}\psi(x)=E\psi(x) $$

And giving the box a width with value $L$, then:

$$ \psi(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{(n+1)\pi}{L}x\right)\quad\forall n\in\mathbb{N} $$

Knowing this, the probability of finding the particle inside an interval $(x,x+dx)$, should be:

$$ |\psi(x)|^{2}dx=\frac{2}{L}\sin^{2}\left(\frac{(n+1)\pi}{L}x\right)dx $$

But $\psi(x)$ looks like a particle with a well-defined momentum:

$$ p=\hbar k=\frac{\hbar(n+1)\pi}{L} $$

So my question is, how can a wave function with a well-defined momentum have a relatively nice uncertainty on position? Shouldn't it present total uncertainty, for example with $|\psi(x)|^{2}$ being a constant over all the space, in order to agree with the Heisenberg's uncertainty principle?

Of course, I know that I'm wrong, but I need to know which is the fact that I'm not correctly understanding .

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  • $\begingroup$ I think, this is easier to see if you take $\psi_+(x)=\exp(ik_nx)$ and $\psi_-(x)=\exp(-ik_nx)$ as your eigenfunction basis (check that those are also solutions of the Schrödinger equation). Then $|\psi_+(x)|^2=|\psi_-(x)|^2=1$ $\endgroup$ – Photon Jan 10 '17 at 19:58
  • $\begingroup$ Momentum of a classical particle bouncing in the box is $\pm p; \ \Delta p = 2p$. $\endgroup$ – Pieter Jan 10 '17 at 20:00
  • $\begingroup$ Although it's not incorrect to use $n+1$ for $n=0,1,2...$ but usually we use $n$ for $n=1,2,3...$ $\endgroup$ – Gert Jan 10 '17 at 20:51
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$$p=\hbar k,$$

is applicable to a free particle. But your particle in a 1D box is bound.

To find momentum for the particle in the 1D box (with infinite potential walls), apply the quantum momentum operator to the wave function:

$$\boxed{\hat{p}_x=-i\hbar\frac{\mathrm{\partial}}{\mathrm{\partial}x}}$$

The expectation value (mean) $\langle p_x \rangle$ for the particle's momentum is actually zero.

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  • $\begingroup$ in order to apply that operator, what should I do? $\left<\psi|\hat{p}|\psi\right>$ $\endgroup$ – Jaime_mc2 Jan 10 '17 at 20:21
  • $\begingroup$ $\left<\psi|\hat{p}|\psi\right>$ gives you the expectation value, yes. en.wikipedia.org/wiki/… $\endgroup$ – Gert Jan 10 '17 at 20:46

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