State Vector vs wave function

Question

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 $$|\psi\rangle=\int dx \;\phi(x)e^{-iEt/\hbar} |E\rangle $$?

Answer 1

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 $$|\psi(t)\rangle=\int dx \;\phi_E(x)e^{-iEt/\hbar} |x\rangle.$$

Answer 2

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.