Is wave function an analogue of probability amplitude or a ket vector from Dirac notation?

Question

The way I was introduced to wavefunctions was in form of Dirac notation: $$\psi(x)=\langle x| \psi \rangle$$ i.e. the probability amplitude of going from state $\lvert \psi \rangle$ to state $\lvert x \rangle$, where state $\lvert x \rangle$ is a member of eigenvectors of some observable, forming a basis. It also makes intuitive sense because the wavefunction peaks at a specific $x$ values which happens to be eigenvalues of the state $\lvert \psi \rangle$. Another example of wavefunction as the probability amplitude is in this notation: $$\lvert \psi \rangle = \sum_i \psi(\textbf{i}) \lvert i \rangle $$ (eigenvector basis expansion of an arbitrary state).

But now, I am being exposed to notation of the sort $X \psi(x)=x \psi(x)$ or $A \psi = \lambda \psi$, which is rather confusing because the wavefunction is being treated like a ket vector or state rather than a probability amplitude. So what is the correct interpretation of wavefunction?

I am considering $\lvert \psi \rangle$ (arbitrary state vector) as being different from $\psi (x)$ (wavefunction).

Answer 1

If i have a vector $\vec{v}$ I can find its components by taking the dot product with basis vectors $$ v_i = \vec{e}_i\cdot \vec{v}\;. $$ I can then write $\vec{v}$ in terms of its components as a column vector. $$ \vec{v} = \left(\begin{array}{c}v_1\\v_2\\\vdots \end{array}\right)\;. $$

The abstract vector notation $\vec{v}$ and the column vector notation describe the same object and ultimately I can extract the same inforamtion from both, but in a given situation one may be much more conveinent than the other.

Now Dirac notation is essentially a variant of abstract vector notation and wavefunction notation can be thought of as analogous to column vector notation (in the $x$ basis). Both the ket $|\psi\rangle$ and the wavefunction $\psi(x)$ contain the full information about the state of the quantum system. I can extract the same information from both (though it may be easier in one notation than the other) and can apply operators to either (although I need to find a representation of the operator to match how I represent the state).

Answer 2

The two notations you have mentioned only differ in their conventions, for example, the first one follows Dirac bra-ket notation $<x|\psi>$ which is the probability of |$\psi$> to go to state |x>.
The second notation $X \psi(x) = x\psi(x)$ is the eigenvalue equation, and says that $x$ is an eigenvalue of the eigenfunction $\psi(x)$ for the operator $X$.
The two here convey different meanings and must not be confused.
Not sure if you have already checked, this explains well. Check: DIRAC’s BRA AND KET NOTATION

Answer 3

It is just a shorthand. Finding the eigenvalues in terms of the wavefunction on real space $X\psi(x)=x\psi(x)$ usually makes more sense when taking expectation values:

$$\left\langle X\right\rangle = \left\langle\psi|X|\psi\right\rangle=\int dx\psi^*(x)x\psi(x)$$

Thus the shorthand for: $X\psi(x)=x\psi(x)$ since the action of $X$ on $\psi(x)$ gives an eigenvalue of $x$ in position space. Since we don't have an explicit expression for the $X$ operator in real space it is easier to derive this expectation via states $|x>$:

$$\left\langle X\right\rangle = \left\langle\psi|X|\psi\right\rangle=\int dydx\left\langle\psi|y\right\rangle\left\langle y|X|x\right\rangle\left\langle x|\psi\right\rangle$$

$$=\int dydx\psi^*(y)x\delta(x-y)\psi(x)=\int dx\psi^*(x)x\psi(x)$$

Using the orthogonality relations: $\left\langle y|x\right\rangle=\delta(x-y)$ where $|x>$ is the set of eigenvectors of the $X$ operator with eigenvalues $x$ and the completeness relation: $\int dx|x\left\rangle\right\langle x|=1$.