Shouldn't any measurement cause all wavefunctions to collapse?

Question

Given the fact that every wavefunction exists everywhere in space, shouldn't a measurement at any location cause all wave functions to collapse since a measurement at any point measures all wavefunctions?

Answer 1

Given the fact that every wavefunction exists everywhere in space, shouldn't a measurement at any location cause all wave functions to collapse since a measurement at any point measures all wavefunctions?

No. Or, alternatively, this question has no answer since it is just a muddle.

Wavefunctions are used to describe microscopic systems, such as atoms or molecules. Wavefunctions provide us with as much information as we could know about the given microscopic system (provided we can solve the governing equations), but wavefunctions are not the microscopic system.

Also, wavefunction "collapse" is not a physical collapse. It describes a change of our knowledge of the given microscopic systems (due to the measurement result being obtained) that we force into our equations "by hand."

For example, a wavefunction $\psi(\vec r)$ might describe the probability amplitude to measure the position of an electron in a hydrogen atom. If you measure the position of that electron in that hydrogen atom, you could measure any value, say, $\vec r_0$. (Any sets of three real numbers are the eigenvalues of the position operator). If this $\vec r_0$ is actually measured (and the probability density that this would happen is $|\psi(\vec r_0)|^2$), then the wavefunction "collapses" to the state $|\vec r_0\rangle$. This just means that, if we would like to further describe the dynamics of the microscopic system, we need to evolve the state $|\vec r_0\rangle$, not the state $\int d^3r \psi(\vec r)|\vec r\rangle$.

So, that specific wavefunction "collapsed." All this means is that our new description of the system will use the prior information. We measured $\vec r_0$ and so we know the state right now is $|\vec r_0\rangle$. And we know after an additional amount of time $t$ (if we don't measure anymore) the state will be $e^{-i\hat H t}|\vec r_0\rangle$.

Answer 2

In a comment you clarified your question:

(because the electron wavefuction is everywhere, so isn't not finding the electron at any location in space the same as finding it somewhere specific)?

No, it isn't the same. Realistic position measurements (unlike those in undergraduate QM) have only a small number of possible outcomes, perhaps even as small as two: "found here" and "not found here". The wavefunction collapse is a projection to the eigenspace associated with the measurement outcome. The eigenspace for "not found here" is very large since it includes everywhere else the particle might be, so the projection has correspondingly little effect.

The collapse after a "found here" result will turn the wave function into a narrow spike at the location of the measurement device, but the collapse after "not found here" will just force the wave function to zero at the location of the measurement device, without localizing it elsewhere.

Answer 3

What does "collapse" mean? Depending on your formalism, yes, any measurement of anything will "collapse" the wave function of the universe into an eigenspace of that measurement. But surely that eigenspace is considerably more than one-dimensional.