does collapse EVER result in classical states?

[gib65]

The standard theory of wavefunction collapse describes particles existing in states of superposition, and then when they are measured they collapse into a classical state. But I think this is a little misleading. I could be wrong, so I pose this as a question. Isn't measurement never perfectly precise? Which means that whatever you're measuring (position, momentum, energy, time, etc.) is going to have some margin of error. The uncertainty principle tells us that this is not a shortcoming of our measuring instruments but with the nature of the superpositions states themselves. So if you want to measure position, you're necessarily going to have to accept the uncertainty in momentum. But when do we ever measure position precisely such that momentum is going to be infinitely uncertain? This leads me to wonder if whatever it is we're measuring is always going to have some degree of uncertainty to it (and this uncertainty is inherent in the thing we're measuring), and so if measurement collapses the wavefunction, the collapse is never going to achieve classical states - it will only approximate them. Is this sound reasoning?

[Tom Mattson]

I don't know what you mean by "classical state". To me that suggests that the motion of a particle satisfies Newtonian/Maxwellian equations. If that's what you mean than no, wavefunctions cannot collapse to such states, even in principle. What they collapse to is an eigenstate of the observable being measured. Pure eigenstates of QM operators are still quantum states.

[Wormwood]

I might have misread the OP and if so I apologize, but it seems as if you think that a phenomenon must be measured in a proper sense to collapse the wave function. A phenomenon must simply be observed, not measured; though measurement is a type of observation. Does that help at all?

 

Also Tom, I think by "classical states" the OP just means realized potential states; or states where the wave function has been collapsed into one definite phenomenon.

 

Again, if I have misread anything I apologize.

[gib65]

I don't know what you mean by "classical state". To me that suggests that the motion of a particle satisfies Newtonian/Maxwellian equations.

 

That is exactly what I meant.

 

it seems as if you think that a phenomenon must be measured in a proper sense to collapse the wave function. A phenomenon must simply be observed' date=' not measured

[/quote']

 

Well, that's one interpretation (the Copenhagen one I believe). Both observation and measurement are specific examples of conditions under which collapse will occur, and as you said they are, for the most part, the same thing. In any case, what I'm asking is, whatever the conditions for collapse ultimately are, does collapse ever result in "classical states" as Tom correctly interpreted it (Newtonian/Maxwellian).

[Tom Mattson]

That is exactly what I meant.

 

In that case, no wavefunctions do not ever satisfy the equations of classical physics. However the expectation values of operators can satisfy them (more on this later). Here's what does happen.

 

You've got a wavefunction that is a normalized superposition of the eigenstates of the observable you're measuring. Let's say we're measuring some generic observable with operator [math]\hat{A}[/math] and let's say that the eigenstates of [math]\hat{A}[/math] are denoted by [math]\phi_n(\vec{x},t)[/math] with eigenvalue [math]a_n[/math].

 

So we have the following eigenvalue equation.

 

[math]\hat{A}\phi_n(\vec{x},t)=a_n\phi_n(\vec{x},t)[/math]

 

And we can write down the wavefunction [math]\psi(\vec{x},t)[/math] of our system as a linear combination of the basis eigenstates as follows.

 

[math]\psi(\vec{x},t)=\sum_{n=1}^{\infty}c_n\phi_n(\vec{x},t)[/math],

 

where [math]\sum_{n=0}^{\infty}|c_n|^2=1[/math]. Now it's important to remember that not only is [math]\psi(\vec{x},t)[/math] a quantum state (that is, it satsifies the Schrodinger equation), but so are all of the [math]\phi_n(\vec{x},t)[/math]. When you make a measurement of observable [math]\hat{A}[/math], all you do is force the system into one of the eigenstates. Let's say the system jumps into the [math]k^{th}[/math] one. So instead of finding our system in the above linear combination, we find that our system jumps: [math]\psi(\vec{x},t)\longrightarrow\phi_k(\vec{x},t)[/math], and the probability for doing so is [math]|c_k|^2[/math].

 

So the system is in a pure eigenstate of [math]\hat{A}[/math]. But it's still a quantum state.