In Dirac's state vector notation, the position representation is given by :
$$ |\psi\rangle = \int d^3r\;\psi(\mathbf{r})|\mathbf{r}\rangle$$ My questions:

  1. Is the State Vector Different from the wave function?
  2. Can there be an energy representation of the state vector,
  3. Why do people write $|\psi(t)\rangle$ if the state vector is abstract?
  4. Suppose the wavefuction at the time $t$ is some $\phi(x)e^{-iEt/\hbar}$. Such a state is indeed an energy eigenstate, can one then write $$|\psi\rangle=\int dx \;\phi(x)e^{-iEt/\hbar} |E\rangle $$?
  • 1
    $\begingroup$ Your first question sounds like a duplicate of this one. $\endgroup$
    – Sandejo
    Dec 8, 2021 at 2:22

2 Answers 2

  1. Yes, the state vector $|\psi\rangle$ and the wave functions $\psi(r)$ are different things. But they are isomorphic, meaning you can express one in terms of the other and vice versa: $$|\psi\rangle = \int d^3r\;\psi(\mathbf{r})|\mathbf{r}\rangle$$ or $$\psi(\mathbf{r})=\langle\mathbf{r}|\psi\rangle$$
  2. Yes, you can express the state vector in terms of the energy eigenstates: $$|\psi\rangle = \sum_n c_n |E_n\rangle$$ with some coefficients $$c_n=\langle E_n|\psi\rangle$$
  3. I'm not sure if I understand this question correctly.
    The state vector usually varies with time $t$. Therefore we write it as $|\psi(t)\rangle$.
  4. Not quite. You could either write it as a time-dependent abstract state vector $$|\psi(t)\rangle=e^{-iEt/\hbar} |E\rangle$$ or as a time-dependent wave function $$\psi(x,t)=\phi_E(x)e^{-iEt/\hbar}$$ where $|E\rangle$ is an energy eigenstate and $\phi_E(x)$ is the corresponding energy eigenfunction. Both are related by $$|E\rangle=\int dx\;\phi_E(x)|x\rangle$$ which gives you $$|\psi(t)\rangle=\int dx \;\phi_E(x)e^{-iEt/\hbar} |x\rangle.$$

Yes. The state vector is different from the wavefunction. The wavefunction is the set of components of the state vector in a particular basis --- that of the $\hat x$ eigenstates $|x\rangle$.

If the state depends on $t$ then it is not unreasonable to write $$ \psi(x,t)= \langle x|\psi(t)\rangle $$ as $t$ is a parameter not a position like $x$. If the state is an energy eigenstate then $|\psi(t)\rangle = e^{-iEt/\hbar} |\psi(t=0)\rangle$, and $\psi(x,t)= \langle x|\psi(t)\rangle$, so your equation is correct.


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