# State Vector vs wave function [closed]

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 $$|\psi\rangle=\int dx \;\phi(x)e^{-iEt/\hbar} |E\rangle$$?
• Your first question sounds like a duplicate of this one. Dec 8, 2021 at 2:22

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 $$c_n=\langle E_n|\psi\rangle$$
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.