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).


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).

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    $\begingroup$ WAIT! Are you telling me that the continuous variable x of the wavefunction is analogous to the basis vectors (components) of a state vector!!! OMG that makes so much intuitive sense now! A wavefunction contains probability amplitudes for different x values just like a state vector contains probability amplitudes for different components. True, a state vector has a continuous variable spanning from negative infinity to positive infinity while a state vector has finite and discrete components. Does this description make sense @BySymmetry? $\endgroup$ – Eesh Starryn Jul 2 at 14:59
  • $\begingroup$ Essentially yes. I think so. I should add to this a standard caveat to always be careful about applying intution from finite dimension vector spaces to infinite dimensional ones (esspesially when using a contiuously indexed basis). Infinite dimensional spaces have some wierd properties that can surprise you if you blindly treat them the same way as their finite dimensionl conterparts. $\endgroup$ – By Symmetry Jul 2 at 15:12
  • $\begingroup$ Man o man..infinite dimensional spaces having weird properties. oof I can't wait to find them out! Thanks, you just made my day! $\endgroup$ – Eesh Starryn Jul 2 at 15:15
  • $\begingroup$ @EeshStarryn It's really a shame that this idea isn't taught as a matter of routine in classes. Thinking of $\psi_x(x) = \langle x | \psi \rangle$ as a component of the vector $\lvert \psi \rangle$ in the "x basis" is very helpful. You will note that of course we can choose other bases, such as the momentum basis, which produces other wave functions $\psi_p(p) = \langle p | \psi \rangle$. These other wave functions are just different representations of the same quantum state. The $x$ and $p$ wave functions are Fourier transforms of each other... $\endgroup$ – DanielSank Jul 2 at 18:56

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

  • $\begingroup$ I mentioned 2 different notations to show how the wavefunction is being used in different contexts. That's why I'm confused on how to interpret wavefunction and what it is analogous to in Dirac Notation! $\endgroup$ – Eesh Starryn Jul 2 at 13:01

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: $\int dx|x\left\rangle\right\langle x|=1$.


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