# If you measure the energy of a system how does wave function collapse

1. Apr 22, 2014

### cooev769

Hey.

Given that if you measure the energy of a wave function, the wave function must collapse to the eigenstate corresponding to the eigenvalue measured. Does that mean when you measure the energy of a wave function it must collapse the wave function into one of these stationary states?

But then the thing is, if you collapse the wave function to this stationary state, well this stationary state doesn't really do much apart from spin around and oscillate in real and complex space, so would that mean that the wave function is now forever going to be in this stationary state?

Thanks.

2. Apr 22, 2014

### The_Duck

Yes.

Yes, unless you disturb the system somehow in the future. That's why we call them stationary states.

3. Apr 22, 2014

### cooev769

So if you didn't disturb the system again, the system would stay in that stationary state for ever?

4. Apr 22, 2014

### Staff: Mentor

Yes - but that is extremely unlikely.

Indeed in QFT quantum vacuum fluctuations also have an effect.

Thanks
Bill

5. Apr 22, 2014

### cooev769

Sweet, I just mean theoretically, thank you for answering my question. My physics professor said if you measured the position you would get a dirac delta function which would slowly spread out over time, but I guess this isn't the case huh?

6. Apr 22, 2014

### Staff: Mentor

A Dirac Delta function is not a stationary state of the free particle Hamiltonian ie its not an eigenstate of energy - its an eigenstate of position. It spreads for sure.

Thanks
Bill

7. Apr 22, 2014

### cooev769

Yeah not of the hamiltonian but of the position operator. So if you measured the position of the particle the wave function must collapse to the dirac delta function with a definite position. How does this wave function then respond with time after collapsing? So what is the mathematics behind the fact that the wave function then spreads after measuring position.

Cheers Bill

8. Apr 22, 2014

### king vitamin

If you make a measurement at a time t' you will obtain the wavefunction $\psi(x)$, which is necessarily an eigenstate of the measured observable. If you know the energy eigenstates (solutions of the time-independent SE) $\phi_n(x)$, you first compute the expansion coefficients

$$c_n = \int dx \phi_n^*(x)\psi(x)$$

where you assume all states are normalized. Now your state for $t>t'$ will be given by

$$\psi(x,t) = \sum_n c_n \phi_n(x) e^{-iE_n (t-t')/\hbar}.$$

If more than one of the $c_n$ corresponding to different energies are nonzero, the probability density $|\psi|^2$ will have time dependence. This describes the time-evolution of the system after an arbitrary measurement (e.g. position).

9. Apr 22, 2014

### aphirst

My understanding of this has been that a Dirac delta function in the position space would actually be unphysical, since technically speaking there is no mathematical mechanism by which such a function could be said to "spread out". I recall reading in a series of papers which happened to be discussing something to do with de Broglie-Bohm theory, a criticism was raised about position measurements doing something similar to this, but which was countered by pointing out that a more appropriate representation would be an arbitrarily thin Gaussian function. A minor difference, but I believe an important one.

I'm afraid I am unable to locate the discussion or papers, though I expect that delving more into that would be off-topic as per this thread. Though if I have misunderstood their content, someone with more familiarity might be able to find the papers, or just simply point out the issues with what I've said.

UPDATE: I found it on arXiv. On compatibility of Bohmian mechanics with standard quantum mechanics by H. Nikolic (Demystifier here on PF). Again, if I've misunderstood, or perhaps drawn a false analogy with cooev769's question, I'd be keen to know.

Last edited: Apr 22, 2014
10. Apr 22, 2014

### Staff: Mentor

11. Apr 22, 2014

### cooev769

Thank you all for your answers. But I was told that all the time dependence term does when you take it on is causes the stationary state to spin around in complex space. Is this true?

12. Apr 22, 2014

### aphirst

The phase(s) of the decomposition of $\psi$ in the energy basis evolve, yes.

I expect that it is straightforward to show that a Gaussian in position-space would evolve outwards, by considering the Gaussian's representation in Fourier space (the momentum basis is equivalent to the energy basis for a free particle, by my understanding, so you just Fourier transform there and back). I also expect that such a treatment is easy to find either online or in most standard undergraduate texts.

Last edited: Apr 22, 2014
13. Apr 22, 2014

### Staff: Mentor

The key word here is STATIONARY.

A state, when viewed as elements of a vector space, is invariant to phase, but the Schroedinger equation changes more than that - its only stationary states that don't change.

That's because states are not really elements of a vector space - they are actually operators - but beginning texts generally don't tell you that.

What you really need to do is see a PROPER development of QM.

THE book for that is Quantum Mechanics - A Modern Development by Ballentine:
https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

Read the first 3 chapters and all will be clear. Its mathematically advanced, but don't be put off by that, you will be able to get the gist.

Thanks
Bill

Last edited by a moderator: May 6, 2017