So, I'm taking a QM 1 course, and we have reached a point where we used Dirac notation to solve two-level systems more efficiently, but our professor never really bothered to explain it further (he didn't even talk about quantum states or Hilbert spaces, in fact he abuses notation by putting the wave function inside the kets, making it even more confusing).

Now here is the thing, we know about operators, of course, but we have only talked about them (and this is what I find in most textbooks), with regard to their eigenvalue equations and their expectation values. Nowhere have I found a comprehensive explanation of what you get when you use one of these operators on an already given quantum state/wavefunction etc.

Say I have a quantum state in a superposition of two angular momentum eigenstates. What do I get when I apply the Hamiltonian on it? Not computationally, but intuitively, what does that result represent?

Now, for my second question: why aren't operators represented by one matrix? Is it a different matrix, for each of the bases on which the state is represented? How do I know which matrix to use for each operation?

Finally, I do wonder why the matrix is represented by the elements $A_{mn}=\int{\psi_n*\hat{A}\psi_m dx} $ (say, in position space), which are also the elements used in finding the expectation value of a superposition of eigenfunctions of another operator?

P.S.: I'm sorry if these questions seem rather trivial, it's just that our professor just threw this at us completely arbitrarily, and I'm unable to find derivations that make intuitive sense.

  • $\begingroup$ A good help for your last question, and in general on how to understand Dirac notation, representations of operators, etc. can be found on my answer here: physics.stackexchange.com/questions/436877/… $\endgroup$ Commented Nov 18, 2018 at 5:38
  • $\begingroup$ Also, you don't actually apply an operator to a state that is a superposition of different (bases) states. You apply the operators to the bases states (the observable's eigenvectors) themselves to acquire their observable values, i.e. their eigenvalues. In the case of the time-independent Schroedinger equation, the state where the Hamiltonian is being applied to is an eigenstate of the Hamiltonian (so it has a definite energy) i.e. not a superposition of other states. These eigenstates of hamiltonian is then a complete basis upon which your system's state can be written as a superposition of. $\endgroup$ Commented Nov 18, 2018 at 8:17
  • $\begingroup$ Thank you for taking the time to answer! I am aware of what you are saying, that they are not used/applied to general abstract states that are superpositions of other states. However, I am wondering if one would acquire a meaningful or useful answer by applying an operator to a state that is not an eigenstate of that operator. $\endgroup$
    – Alex S.
    Commented Nov 18, 2018 at 16:38

1 Answer 1


!!! I preface here that I am actually also in my first semester of quantum so while I feel confident in my answer if someone more experienced comes along please correct me if I am wrong !!!

Edited From Comments below for correctness

Operators in QM are linear transformations mapping $\mathcal{H} \rightarrow \mathcal{H}$. These operators can then be represents as a matrix as you pointed out. So what this means is if we take an operator and act it on some state we will transform this state in some way. To see this, act the operator $\hat{p}$ on a stationary state of the infinite square well. Since these states are not Eigenstates of $\hat{p}$ you will just transform your state into a new one.

The nice thing comes about with Eigenstates:

$$\hat{Q} \left|\psi\right> = q\left|\psi\right>$$ Here, $\hat{Q}$ is our operator and $q$ is it’s eigenvalue. This is where we gain ‘useful’ information about our state.

As for why aren’t all operators represented by one matrix. That is because as you stated the operator depends on your basis. For example:

In the infinite square well our position operator is $\hat{X}$ and our Eigenstates are the stationary states $$\psi_n(x) = \sqrt{\frac{2}{a}}sin\left(\frac{n\pi}{a}x\right)$$ so to get a matrix representation of $\hat{X}$ we use: $$\hat{X}_{mn}=\left<\psi_m|\hat{x}|\psi_n\right>$$ but for the harmonic oscillator not only is our operator set up differently, but also it’s Eigenstates are different from that of the infinite square well. It’s nice to be able to change basis because we can switch from position space to momentum space when it suits us.

When you say that $$A_{mn}=\left<\psi_m|A|\psi_n\right>=\int\psi_m^*A\psi_n dx$$ is the same as how we calculate expectation values, remember, expectation values are weighted probability’s and the probability density is given by the square modulus of the wave function $|\psi_n(x)|^2=\psi_n^*\psi_n$ (notice that the states are the same). So while $$\left<\psi_m|\hat{Q}|\psi_n\right>$$ and $$\left<\psi_n|\hat{Q}|\psi_n\right>$$ may seem similar, they are different statements.

  • $\begingroup$ Only $\hat{X}_{mn} = \langle \psi_m (x) | x| \psi_n (x) \rangle $ is not quite right. Working in position representation one needs to drop the second $|$ in the middle. If you want to work in operators working on kets then you wrote it in the good way but then the x-dependence should not occur in the ket since this assumes position representation of the state, i.e. $\psi (x) = \langle x | \psi \rangle$. $\endgroup$ Commented Nov 20, 2018 at 10:22
  • $\begingroup$ Also the solution to the infinite welk you mentioned should be $\psi(x)$ instead of $|\psi (x) \rangle $ sinxe the expression on the right is in position representation you cannot equate it to an abstract ket vector on the left. So do not write $|\psi(x)\rangle$ when you mean $\psi(x)$! This certaintly has a different mathematical meaning (see comment above). $\endgroup$ Commented Nov 20, 2018 at 10:25
  • $\begingroup$ And do not forget that the position operator is $\hat{x}= x \hat{I}$ only if one works in position representation. In momentum representation you get a different form. In general when we talk out ket vectors it should just be $\hat{x}$ and no more. $\endgroup$ Commented Nov 20, 2018 at 10:28
  • $\begingroup$ @Dani oof, just to make sure that I understand this in the future, I always thought a Ket was just some vector notation. So if something was a vector then you could just put it in a ket. So from my reasoning I would have assumed that since $ \psi_n(x)=\sqrt{\frac{2}{a}}sin\left(\frac{n\pi}{a}x\right) $ is a square integrable function then $\psi_n(x) \in \mathcal{H}$. Since it belongs to a vector space it can be written as a general vector $\left|\psi_n(x)\right>$. I fully believe that this is not quite right like you stated because I felt I was abusing notation, where did I go wrong though? $\endgroup$ Commented Nov 20, 2018 at 16:52
  • $\begingroup$ No it is not. It is quite a different object. In advanced quantum mechanics one often likes to define $\psi (x)$ as the projection of the ket vector $|\psi \rangle$ (note this ket vector has no position dependence. The idea of ket vectors was to escape from this particular representation. As I said, sometimes one works in momentum representation. In equations we have: $ \psi(x) = \langle x| \psi \rangle $ for the position representation and $ \phi(p_x) = \langle p_x | \psi \rangle $. Therefore $|\psi\rangle $ is a complete abstract object in Hilbert space (note not being math rigorous) $\endgroup$ Commented Nov 20, 2018 at 22:35

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