Collapse of state vector for continuous eigenvalues

In summary, the conversation discusses the concept of "collapse" in quantum mechanics and how it relates to measuring an observable with a continuous spectrum of eigenvalues. It is pointed out that the eigenvectors in this case are not physically realizable states. Instead, the state vector collapses to a probability distribution with a narrow range of values, depending on the precision of the measuring device. However, there is uncertainty in determining the precision of the device, which may affect the calculation of the system's evolution. This highlights the need for a careful analysis of each experimental situation and appropriate modeling.
  • #1
bob900
40
0
1. In the many statements of the QM postulates that I've seen, it says that if you measure an observable (such as position) with a continuous spectrum of eigenvalues, on a state such as

138ce1c0d57f9c6da094402128085245.png


then the result will be one of the eigenvalues x, and the state vector will collapse to the eigenvector |x>.

However, for say the position operator, the eigenvectors are the dirac delta functions δ(x-x') and they do not represent physically realizable states (since they can't be normalized). So what does the state vector actually collapse to?

2. In practice there will always be experimental uncertainty Δ due to the resolution of your measuring apparatus, etc. So if you measure some position x, with an uncertainty +/- Δ, does the state vector still collapse to an eigenvector corresponding to some particular eigenvalue within the [x-Δ, x+Δ] range? Or does it collapse perhaps to some (of infinitely many) states ψ(z) such that for z outside of the [x-Δ, x+Δ] range |ψ(z)|^2 = 0, but inside that range the probability amplitudes are equal for all z?
 
Physics news on Phys.org
  • #2
1. I'm not a supporter of the collapse proposal.
2. Point 1 of yours indeed contradicts the first axiom which excludes distributions as physical states.
3. For continuous spectrum, one tries to mend it to the other axioms and resorts to probability distributions with interval [itex] \delta [/itex] in the parameter space of the (real) continuous spectrum.

For pure states:

[tex]|\psi_{new}\rangle = \frac{P_{\delta}|\psi_{old}\rangle}{\langle \psi_{old} |P_{\delta}|\psi_{old}\rangle} [/tex]

[tex] P_{\delta} = \int_{a_{0} - \frac{\delta}{2}}^{a_{0} + \frac{\delta}{2}} d\alpha |\alpha\rangle \langle \alpha| [/tex]
 
  • #3
dextercioby said:
2. Point 1 of yours indeed contradicts the first axiom which excludes distributions as physical states.
3. For continuous spectrum, one tries to mend it to the other axioms and resorts to probability distributions with interval [itex] \delta [/itex] in the parameter space of the (real) continuous spectrum.

For pure states:

[tex]|\psi_{new}\rangle = \frac{P_{\delta}|\psi_{old}\rangle}{\langle \psi_{old} |P_{\delta}|\psi_{old}\rangle} [/tex]

[tex] P_{\delta} = \int_{a_{0} - \frac{\delta}{2}}^{a_{0} + \frac{\delta}{2}} d\alpha |\alpha\rangle \langle \alpha| [/tex]

I think this is roughly equivalent to what a book I have (Griffiths Intro to QM) says :

"In the case of continuous spectra the collapse is to a narrow range about the measured value, depending on the precision of the measuring device".

But how do you ever know the 'precision' of your measuring device and how does it enter your practical calculations? Let's say I make your measurement device really imprecise - I measure a particle's position by having a line of very large (say 10 meter long) detector plates and recording which plate the particle strikes. That will measure position (along one direction) to the accuracy of 10 meters. Once I obtain my result and want to then calculate the evolution of the system further, I will then use, like you stated :

[tex]|\psi_{new}\rangle = \frac{P_{\delta}|\psi_{old}\rangle}{\langle \psi_{old} |P_{\delta}|\psi_{old}\rangle}[/tex]

with δ=10m, as my new initial state.

But it could be said that I have really measured it to more accuracy - perhaps there is some investigation that can be performed on the plate (looking at how its internal structure, at say the molecular level, was affected), that will allow me to find out more precisely where the particle struck, say down to the millimeter? Then I would have to use δ=10^-3m, which will produce quite different predictions than had I used δ=10m!

So it seems that for me to calculate a correct result, I would somehow have to know the maximum precision of my measuring device. How is that possible, or am I missing something basic here?
 
  • #4
bob900 said:
[...] or am I missing something basic here?
Yes. But hey, don't feel bad -- everyone who uses the notion of "collapse" to try and explain QM is also missing that same "something basic"...

Get hold of Ballentine's text "QM -- A Modern Development" and study section 9.2 (and possibly some earlier chapters if sect 9.2 doesn't make sense). The basic idea is that realistic measurement involves establishing a correlation between the system's initial state and the apparatus' final state, via an interaction. The reason this emphasis is less visible in much of introductory QM material is that you can't go very far with this in general form. Rather, one must analyze each experimental situation carefully and model it appropriately. Typically, that needs a fair bit of work for every situation.
 

Related to Collapse of state vector for continuous eigenvalues

1. What is the collapse of state vector for continuous eigenvalues?

The collapse of state vector for continuous eigenvalues is a phenomenon in quantum mechanics where the act of measuring a continuous observable (such as position or momentum) causes the state of a quantum system to "collapse" into a specific state or eigenstate.

2. Why does the collapse of state vector only occur for continuous eigenvalues?

The collapse of state vector only occurs for continuous eigenvalues because discrete eigenvalues (such as energy levels) do not require a continuous measurement process. The value of a discrete eigenstate can be determined through a single measurement, while continuous eigenvalues require multiple measurements to determine the exact value.

3. How does the collapse of state vector affect the uncertainty principle?

The collapse of state vector does not affect the uncertainty principle. The uncertainty principle states that the more precisely one observable is measured, the less precisely the other observable can be known. The collapse of state vector does not change this principle, but rather reveals the state of the system after the measurement has been made.

4. Is the collapse of state vector a physical or mathematical phenomenon?

The collapse of state vector is a mathematical phenomenon that describes the behavior of quantum systems. It is not a physical event, but rather a mathematical representation of how a quantum system evolves when measured.

5. Can the collapse of state vector be observed in real experiments?

The collapse of state vector cannot be directly observed in real experiments, as it is a mathematical concept. However, its effects can be observed through experiments that measure continuous observables in quantum systems.

Similar threads

  • Quantum Physics
Replies
2
Views
1K
Replies
1
Views
782
  • Quantum Physics
Replies
4
Views
786
  • Quantum Physics
Replies
24
Views
1K
Replies
3
Views
924
  • Quantum Physics
Replies
6
Views
2K
Replies
32
Views
2K
Replies
1
Views
850
  • Quantum Physics
Replies
9
Views
998
Replies
20
Views
3K
Back
Top