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I've started studying quantum physics and I'm getting confused on some key concepts. What I understand so far: we have a mathematical object called a wave function from which we can extract quantities properties or observables to describe our quantum system. To get these quantities, we will apply an operator on our wave function which is Hermitian and linear like $\hat X$ or $\hat P$ in 1D. I don't understand what the following statement means:$$$$ 'The eigenvalues of the operator represent the possible results of carrying out a measurement of the corresponding quantity. '

what does any of this have to do with linear algebra and eigenvalues? Why do we forget the wave function and start using an 'eigenfunction' once a measurement is made? Also what do these operators return? For example, if $\psi (x) = Ae^{ikx}$ then $\hat P \psi = -i \hbar \frac {d \psi}{dx} = \hbar k \psi$. $\hat P $ is supposedly the momentum operator but returns the momentum times the wave function why is that?

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The eigenfunctions of a Hermitian operator form a complete basis, so that any function can be expanded in terms of these eigenfunctions as $$ \psi = \sum_nc_n\phi_n \text{ where } \hat{O}\phi_n = o_n\phi_n. $$ If the wave function is one of the operator eigenfunctions, e.g. $\phi_k$, then measuring the quantity produces the corresponding eigenvalue, $o_k$. However, if the wave function is not an eigenfunction, i.e., it is a superposition of several eigenstates, then we will measure value $o_k$ with probability $|c_k|^2$. Note that in practice we do multiple measurements on identical objects, so $|c_k|^2$ stands for the fraction of those measurements where the value $o_k$ is observed.

It is easy to see that, if the eigenfunctions of $\hat{O}$ are chosen to be orthogonal and normalized, then $$ \langle \hat{O}\rangle = \int dx \psi^*\hat{O}\psi = \sum_n o_n|c_n|^2 $$

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'The eigenvalues of the operator represent the possible results of carrying out a measurement of the corresponding quantity' is just another way of saying that an operator is the mathematical way of writing the experiment we do on the state and the eigenvalue is the result or the observable we get from that experiment...For example when we apply the momentum operator on a particle, we get a matrix and then we have to "factor out" (normalize) an eigenvalue which turns out to be the momentum of the particle...You said that you get back the momentum multiplied by the state $\psi$ but don't think about it that way...when you actually apply the momentum operator on $\psi$ you get a matrix and then you have to 'factor out' a value so that the other matrix is $\psi$...In the end you should have this-

$(Momentum- operator)\psi=(eigenvalue)\psi$

You might say that the eigenvalue and the momentum operator will be equal but if you dig into the subject you'll see that they don't...(It is a bit complicated)...So the result doesn't give you the momentum of the particle...the eigenvalue or the factor that you 'factored out' does...

Hope you got your answer...

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You ask why we forget the wavefunction,namely the state of the quantum object when a measurement is applied.For this,I want to point out that what we really care about is the measurable quantities in a measurement,namely what value can we get through this measurement and what the probability of getting that.We don't care the state of the system(wavefunction) because it's just a mathematical tool that describes our physics system,and you can't measure the wavefunction of a system(I mean a really physical measurement).And that statement:The eigenvalues of the operator represent the possible results of carrying out a measurement of the corresponding quantity.It's due to one of fundamental assumptions in quantum physics,it's saying that when applying a measurement on system,the state of the system will collapse into a eigenstate of this measurement,meanwhile we get the eigenvalue corresponding to the eigenstate.

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we have a mathematical object called a wave function from which we can extract quantities properties or observables to describe our quantum system. To get these quantities, we will apply an operator on our wave function which is Hermitian and linear like $\hat X$ or $\hat P$ in 1D.

No, this is not the case. It is true that the wavefunction (or more generally the state vector) tells us information about our system. More specifically, it can tell us the probability of measuring the value of some observable. However, it is not the case that this information is obtained by applying operators to the state vector (unless the operator is the sum of the projection operators (unity for a complete basis), as shown below).

Instead we do the following. The state vector $|\psi\rangle$ is an abstract mathematical object. Our observables represented by Hermitian operators have real eigenvalues and permit construction of complete, orthonormal eigenbases. When we take an observable $A$ and project $|\psi\rangle$ onto its eigenbasis $\{|a_i\rangle:i=1,2,3,\dots\}$ so that

$$|\psi\rangle=\sum_i|a_i\rangle\langle a_i|\psi\rangle$$

Then the values $|\langle a_i|\psi\rangle|^2$ tell us the probability of getting a result of eigenvalue $a_i$ when we measure $A$ (assuming you normalized the above sum).

And this gets to your statement

The eigenvalues of the operator represent the possible results of carrying out a measurement of the corresponding quantity.

This is a postulate of Quantum Mechanics. If we have a Hermitian operator associated with an observable, then we can only obtain measurement results equal to eigenvalues of the operator.

The above also addresses other concerns

what does any of this have to do with linear algebra and eigenvalues?

I hope it is a little more obvious now. We are working with expressing vectors in various eigenbases.

what do these operators return? For example, if $ψ(x)=Ae^{ikx}$ then $\hat Pψ=−iℏ\frac{dψ}{dx}=ℏkψ$. $\hat P$ is supposedly the momentum operator but returns the momentum times the wave function why is that?

First, it returns the momentum times the wave function in this specific instance because you started with a momentum eigenstate already. In general you would not get just a constant times the wave function. And second, like I mentioned above, applying the operator to the state vector is not how you go about obtaining the possible momentum measurements. (I suppose this does actually apply for eigenstates since you do get back the eigenvalue, but other than that this is not the case).

Why do we forget the wave function and start using an 'eigenfunction' once a measurement is made?

This is actually a separate Quantum Mechanics Postulate. If you get some eigenvalue for a measurement then the state vector right after measurement is the eigenvector corresponding to that eigenvalue (or in the case of degeneracy the state vector is in the eigenspace corresponding to that eigenvalue).


So hopefully you see now that applying an operator on your state vector doesn't then give you the observable corresponding to system (in general your system is not in an eigenstate of an observable, so you would not expect to get just some constant back out).

But what are operators useful for then? One use is described above and is probably the most important part. But there are instances where actually applying operators to states is useful. One such example is expected values. Let's say I wanted to know the expected value of measurements of $A$. Well according to the mathematics of probability this would be equal to

\begin{align} \langle A\rangle &= \sum_i|\langle a_i|\psi\rangle|^2a_i\\ &= \sum_i(\langle\psi|a_i\rangle\langle a_i|\psi\rangle) a_i\\ &= \sum_i\langle\psi|a_i|a_i\rangle\langle a_i|\psi\rangle\\ &= \sum_i\langle\psi|A|a_i\rangle\langle a_i|\psi\rangle\\ &= \langle\psi|A\left(\sum_i|a_i\rangle\langle a_i|\right)|\psi\rangle\\ &= \langle\psi|A|\psi\rangle \end{align}

Another example where applying operators to the state vector is used is in the propagator of the system, which tells us how the quantum state evolves over time. We usually denote this as $$|\Psi(t)\rangle=\hat U|\Psi(0)\rangle$$ where $\hat U=\exp(-i\hat Ht/\hbar)$ is the propagator with the Hamiltonian of the system $\hat H$.

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The algrebra and related discussion is necessary. And if you are "up" on that and work with it a while, it can be sufficient.

I like to put some experiments with it to have some idea of how this looks in the lab. The Stern-Gerlach experiment is very instructive. In this experiment, silver atoms are streamed through a magnetic field with a strong gradient.

enter image description here

The blue lines are beams of silver atoms. They are prepared in a chamber where they are heated to produce a beam. Because they are produced this way they come out in random conditions. The beam splits into two, one spin up and the other spin down. The magnets act as a spin-operator that splits the atoms into spin up and spin down. The spin of the atoms is associated with a magnetic field. When it is parallel to the magnetic field in the lab magnets, it goes one way. When it is anti-parallel it goes the other.

There are a bunch of interesting experiments you can do with equipment like this. For example, if you take the top beam in this setup and put it through another similar device, you can turn the relative angle and see interesting things.

If the two are parallel with the same orientation, then the second device sees only the upper beam, no lower beam. The first device has pulled out only the spin up atoms. The second gets only spin up and so it does not find any spin down.

But if you turn the second device 90 degrees to the first, you get two beams again. That is, spin-up is not in an eigenstate of spin-sideways. And if you slowly turn the second device through various angles, you get changing intensity of the two beams from all one, through a mixture of some of each, to all the other.

There are other keen experiments. You should search around on the net for some of the foundational experiments. They make interesting reading. Some of them are even fairly easily reproducible with university lab equipment.

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