# What state does a particle take on when we measure its momentum?

1. Jul 23, 2013

### andrewkirk

According to the QM postulates, when we measure momentum, we get an eigenvalue of the momentum operator and the system takes on a state that is the corresponding eigenket.

Now, as I understand it, the eigenkets of the momentum operator are all Dirac delta functions corresponding to plane waves. A particle whose state vector is like that has a uniform probability of being anywhere at all in the universe. My understanding from Shankar is that no system actually has a plane wave state vector, and that these are just devices for helping us calculate and think about simple systems.

Shankar (2nd ed. p138) seems to imply that the state immediately after we take a momentum measurement is a vector whose representation in the momentum space is a very sharply peaked Gaussian-style curve centred around the result that we got. The representation in the coordinate space is then a broader Gaussian.

The trouble is that interpretation doesn't seem consistent with the postulate, which says that the state changes to an eigenket of the momentum operator. No matter how narrow and sharply peaked the Gaussian, it is still not an eigenket of that operator.

How can this puzzle be resolved?

2. Jul 23, 2013

### wotanub

It only becomes an eigenstate when we measure the state with absolute certainty.

There is always going to be some uncertainty in your measurement.

3. Jul 23, 2013

### neerajareen

First we must try and clarify a few things. There are 2 sets of basis that we can work one. One is position basis where a definite position is a delta function of position. The other is the momentum basis where a definite momentum is a delta function of momentum. As you correctly pointed out, the eigenstate of momentum is a complex plane wave but this is only true in the position basis.

Now we come to the idea of measurement. In reality, we never measure anything so precisely such that it's wave function is a delta function. The best we can do is come close (know approximately where the particle is or approximately what momentum it has). Therefore in reality, most wave functions tend to be gaussians. Now if we use a significantly high vertical stretch and significantly high horizontal compression, this Gaussian starts to look a lot like a delta function.

Everything is consistent. It's just that you can use different basis to express physics ( position or momentum basis ) plane waves in one bases become delta functions in other. Gaussian are approximate delta functions due to the imprecisions in experiment

4. Jul 24, 2013

### andrewkirk

Thank you for the replies. I have a couple of follow-up questions:

1. Would it be correct to say that in practice, where the eigenvalue spectrum of the operator Ω corresponding to the quantity ω we are measuring is continuous, the system state changes, upon measurement, to a ket ψ whose representation in the Ω basis is a Gaussian centred on the value of ω our measurement gave us?

2. Is the new state ψ an eigenket of any operator we can readily describe, or do we need to let go of the eigenket idea when talking about practical measurements over a continuous spectrum?

5. Jul 24, 2013

### neerajareen

1) That is correct. When you measure a certain observable Ω, the eigenvalue is ω and the eigenvector is ψ in the appropriate basis. It will yield a gaussian in the basis of all the eigenvectors of Ω.

2) Experimentally, measurements are very tricky. For example when when we measure position of particle, experimentally we get a gaussian. But a gaussian is not an eigenfunction of position. The delta function is. The crux of the matter is that we cannot measure anything to infinite accuracy.

Hope that helps

6. Jul 24, 2013

### StarsRuler

Not always: when you measure a continuous observable, the wavefunction collapses to a function that it is the original function in values inside of uncertainty range of the measure, and 0 in the rest, only that this new function has a norm that it is not 1 and we must normalize it. It is not necessarily a gaussian. You can´t measure a continuous observable with absolute accuracy, thus in practice it never collapses really to a delta function or a non normalizable function in any measurement.

7. Jul 24, 2013

### kith

If you perform a measurement you get an eigenstate of the corresponding operator. If you perform a measurement and you get gaussian distributions of momentum eigenstates as final states, you simply haven't measured the operator P but a slightly different operator PΔp.

I've described a possible position measurement with finite resolution here.

8. Jul 24, 2013

### Jano L.

This is a good question. Similar question about the eigen-vector of the position operator bothered me too. Two things are important:

1) the projection/reduction postulate is not a necessary part of the rest of the mathematical theory and is problematic also for other reasons than the one above (see Ballentine's book Quantum Mechanics, A Modern Development, chap. 9). One does not have to stick to this postulate.

2) the objects like $e^{ipx/\hbar}$, $\delta(x-x_0)$ do not belong to the set of $\psi$ functions allowed by the probability interpretation, since they are not normalizable. In other words, there is no $\psi$ that both obeys the Born interpretation of $|\psi|^2$ and represents a state with definite position or momentum.

9. Jul 24, 2013

### dextercioby

One ascribes physical significance only to (normalized) vectors in the Hilbert space, so there's a need to reformulate the postulate of von Neumann and write it in terms of an 'interval' (but genuine) projector, exactly what Kith said: measurements with finite resolution for any observable with an operator with continuous spectrum (as opposed to pure point).

10. Jul 25, 2013

### StarsRuler

The projection postulate is problematic for some interpretations of QM, not for QM. Of course, there is interpretations with no collapse like Many Worlds (maybe more a slighly different from a simple interpretation) or even Relational Quantum Mechanics somewhat, but at least in the case of MWI, there is some problems with the Born Rule inter alia.