I've started taking an advanced quantum mechanics course and I'm having some trouble understanding some of the concepts related to state space representation. As I understand, $|\psi\rangle\in\mathcal{H}$, where $\mathcal{H}$ is the Hilbert state space, is a general representation of the wave function of a system. It is a vector that, in itself, is independent of the basis you choose to represent your wave function in. The choice of basis leads to the different representations of state space.

So far, the two representations that have been explained to me in class are:

  • the position representation, that you obtain like this: $\langle \vec{r} |\psi\rangle = \psi(\vec{r})$.
  • the momentum representation, that you obtain like this: $\langle \vec{p} |\phi\rangle = \phi(\vec{p})$.

Assuming I got all that right, here comes my question: do all representations hold the same information? I was told that the ket-bra notation is more general than, say, the position representation, and that it would enable us to include information about spin. That suggests that there is some information about the state space that is lost when choosing a particular representation, but that does not make sense if the basis for the position representation is a basis for the whole state space. We should be able to express everything in it, right?

Thanks in advance, and please let me know if there is any way I can improve my question :-).

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    $\begingroup$ This "ket-bra-myth" is a myth. Kets are vectors and bras are (continuous) linear functionals. It is just notation, sometimes useful, sometimes not. Yes, all representations are equivalent. I don't understand the issue with spin. $\endgroup$ Sep 15, 2023 at 18:48

1 Answer 1


It is true that all representations are equivalent.

The problem mentioned with reference to spin is probably just that some physical systems have more degrees of freedom than just a position or a momentum vector. An electron, for example, has an intrinsic angular momentum known as spin in addition to its regular momentum degrees of freedom. This just means that the bra/ket must incorporate information about all of the degrees of freedom and so, too, must the wavefunction. One may write things such as $$\psi(\vec{p};\lambda)=\langle \vec{p};\lambda|\psi\rangle$$ to represent a state with momentum $\vec{p}$ and spin $\lambda$ ($\lambda=\pm 1/2$ for an electron).

In all, there is nothing wrong with the representations you know so far, it is just that they may need to become richer when more degrees of freedom are involved.


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