According to Griffiths, there is a general state vector $|s(t)\rangle$ that encodes the state of the system. He also says that we take $\Psi(x, \ t) \ = \ \langle x | s(t) \rangle$. Would then mean that:

$$\Psi(x, \ t) \ = \ \displaystyle\int \ \delta(y \ - \ x) s(t) \ \text{dy}?$$

Also, Griffiths then says that the functions $\Psi$ for the wavefunction, $\Phi$ for the momentum wavefunction, and the collection ${c_n}$ for discrete energy expansion coefficients are all ways of expressing the same function:

$$\Psi(x, \ t) \ = \ \displaystyle\int \Psi(y, \ t)\delta(x \ - \ y)\text{dy} \ = \ \displaystyle\int \Phi(p, \ t) \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\text{dp} \ = \ \displaystyle\sum \ c_n e^{-iE_nt/\hbar} \psi_n(x)$$

I'm having a bit of disconnect. Why does he suddenly start talking about $\Psi(x, \ t)$ in the position space, when he was just discussing how $s(t)$ is a general construction that gives us the position/momentum wavefunctions when expanded in a particular basis? Is $\Psi(x, \ t)$ equivalent to $s(t)$? If not, where does $s(t)$ fit into the picture? I feel like Griffiths is just taking the general state vector to be defined with respect to position, as he seems to do the same thing when he outlines the general statistical interpretation (he says that for a particle in state $\Psi(x, \ t)$, we take the inner product with an eigenstate of some observable to get the probability of getting its associated eigenvalue upon measurement).

I guess the question that would sum this all up is: does Griffiths choose to express the general state vector in terms of the position space?

  • $\begingroup$ I can't comment in detail, and I don't have Griffiths, but his notation is probably throwing you off. I think that there is no function $s(t)$. He probably means a state vector that can change in time. More like $|s>(t)$, but that looks strange. Think about that until someone corrects me or expands on this. State vectors like |s> are abstract. They don't acquire a manifestation until projected onto some space that we can measure, in this case, real position space. $\endgroup$
    – garyp
    Jul 11, 2019 at 2:27

1 Answer 1


does Griffiths choose to express the general state vector in terms of the position space?

Yes. He also projects it onto the momentum basis can calls it $\Phi(p,t)$.

The point is that $s(t)$ can tell us anything we want to know about our system, but it's only useful once we project it onto some basis so we know the probabilities of measuring the system of being in one of those basis states.

Project $s(t)$ onto the position basis, we get $\Psi(x,t)$. It tells us the probability of finding the particle from position $x$ to $x+\text dx$. Project $s(t)$ onto the momentum basis, we get $\Phi(p,t)$. It tells us the probability of measuring the particle to have a momentum ranging from $p$ to $p+\text dp$.

You ask if all of these things are equivalent. I would say yes and no. They give us different information, but they all describe the same system. I like to think of it as we have this abstract thing $s(t)$ that we can only describe in terms of it's shadows (projections). Plato's cave allegory comes to mind.

Links for more reading about basis vectors in QM:

What are basis vectors?

Basis vectors in their own basis

More basis stuff

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Jul 11, 2019 at 16:41

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