alexepascual
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svnaras said:Like all beginners to QM I'm really confused about the measurement operation. I understand that measurement is simply a dot product with an "operator" and the result is one of the operator's eigenvalues.
Now my question is what exactly is an operator? If someone could explain what physical entity is an operator in the following situations that would help me understand this better.
1. A particle moving at some velocity hits a wall/detector. Till it hits the detector its position is described by a combination of position eigenstates. Once it hits the detector its position becomes a single eigenvalue. Is the wall an operator here?
2. A particle moves through an SG apparatus. Till it passes through the apparatus its spin state is a combination of two eigenstates (in some axis). Once it moves through the magnetic field its spin state becomes one of the eigenstates. Is the magnetic field an operator here?
To all posters on this thread:
I thing svnaras has asked very simple questions and you all have failed to give a simple answer. svnaras must be even more confused than when he first started the thread.
I tried to provide an answer in two oportunities but had some problems submitting my post and lost the text in both oportunities. So I am a little frustrated about these technical dificulties and feel somewhat reluctand to try again.
The questions svnaras asked can be answered according to the orthodox interpretation concentrating more on how the math works than anything else. I'll try to give an answer even shorter (and necessarily incomplete) than my previous answers:
(1) An operator that represents an observable (which can be represented as a matrix for discreet variables) is used to obtain a weighted average of the possible values that the measurement may yield. The weights are the probabilities which are encoded in the state vector. To get the average you multiply (inner product) the matrix by the state vector on both sides. This gives you a single (real) number.
(2)An act of measurement is not described using the observable's operator. It is usually described using a projector. The projector in its simplest expression is a matrix with all zeros except one element in the diagonal that has a 1. To get the final state, you multiply the state vector by the projection operator and get the the new state vector. If you want to get the actual value (momentum, position etc.) that this state vector represents, you can use the procedure I described in (1) using the new state vector.
(3) The wall is not an operator. The collapse that happened at the wall can be represented using the projector operator.
(4) In SG the magnetic field only allows you to select a particular basis. Each orientation in space is a different basis (frame of reference). The magneti field does not collapse the state vector to one of the eigenvalues (you can re-combine the beams). See Feynman Lectures on Physics volume III. If you put an obstacle in one of the paths, then this is a measurement that can be represented using a projection operator. If both pathsare open, instead of a projector, you got an identity matrix. If you put an obstacle in the lower path, then you this kills the lower eigenstate. So you put a zero in the diagonal's lower right.
So, going back to your question, the magnetic field could be seen as a change of basis unitary operator.