# Violation of the laws of quantum physics?

1. Nov 14, 2008

### kasse

If I want to calculate <x> of an electron that is in a state described by $$\psi (x,t)=\frac{1}{\sqrt{2}}[\psi_{0}(x,t) + \psi_{1}(x,t)]]$$ where $$\psi_{0}$$ and $$\psi_{1}$$ are the two lowest energy states of an electron in a one dimensional box, can I then simply calculate <x> for $$\frac{1}{\sqrt{2}}\psi_{0}$$ and $$\frac{1}{\sqrt{2}}\psi_{1}$$ alone and add them?

Is it even possible for an electron to have this energy? Doesn't it violate the laws of quantum physics? I find it strange that this was an exam problem in quantum physics at my university recently.

Last edited: Nov 14, 2008
2. Nov 14, 2008

### dabi

You know, what <x> means?
It is $$\langle x\rangle=\frac12\int_{-\infty}^{\infty}dx x (\psi_0^2+\psi_1^2+2\psi_0\psi_1)$$.
Indeed, $$\int_{-\infty}^{\infty}dx 2\psi_0\psi_1=0$$
but $$\int_{-\infty}^{\infty}dx 2x\psi_0\psi_1\neq0$$.
So you can't calculate <x> for $\psi_0$ and $\psi_1$ separately.
Sure, it's a legal state in quantum physics.

http://www.shareapic.net/content.php?id=12668640&owner=dabi [Broken]

Last edited by a moderator: May 3, 2017
3. Nov 14, 2008

### kasse

The energy must still be either E0 or E1, right?

4. Nov 14, 2008

### dabi

If you measure the energy, you will find either E0 or E1. And that is the special property of quantum mechanics, that the probability is for either case 50%.
http://www.shareapic.net/content.php?id=12668333&owner=dabi

5. Nov 14, 2008

### kasse

But it's not possible to measure E with certainty, right?

And the $$\frac{1}{\sqrt{2}}$$ is there just to normalise the probability function?

6. Nov 14, 2008

### borgwal

Given your other post about <p>, first read your textbook!

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