Relations between Dirac-notation representation of wave function and wave function in a particular basis

My lecturer has emphasised on a number of occasions that:

$$\ |\psi \rangle \neq | \psi(p) \rangle \label{a}\tag{1}$$

since $$\psi(p) = \langle p| \psi \rangle = \int \langle p |r \rangle \langle r | \psi \rangle dr. \label{b}\tag{2}$$

However now I am trying to solve a time Independent perturbation problem whereby I am told that the first order perturbation theory is:

$$-\frac{i}{\hbar}\int_{0}^{t}\left<\psi_{b}(r)|V(r,t)|\psi_{a}(r)\right>e^{i\omega_{0}t}dt$$

I am told that $$\psi_n(\phi) = \frac{1}{\sqrt{2 \pi}} e^{in\phi}$$ and am trying to see which transitions are allowed meaning that I have to simply determine for what values of n and n' will ensure the matrix element $$\left<\psi_{n'}(r)|V(r,t)|\psi_{n}(r)\right> \label{c}\tag{3}$$ is non-zero.

However the problem is that my lecturer has stressed (1),(2) and so I am unsure how to represent $$\psi_n(\phi) = \frac{1}{\sqrt{2 \pi}} e^{in\phi}$$ in the form $$\langle \psi_{n'}(r) |$$ and $$| \psi_n(r) \rangle$$ as required for (3).

• There's two ways of looking at this. Either it's an abuse of notation, or your instructor is being unnecessarily pedantic. I favor the former. There's nothing more frustrating than to run across ambiguous or undefined notation. It stops you in your tracks. Dec 26, 2020 at 4:23

Strictly speaking, if one insists that $$|\psi\rangle \neq |\psi(p)\rangle$$ then one should also insist that $$|\psi\rangle \neq |\psi(r)\rangle .$$ In words, the state vector $$|\psi\rangle$$ represents an entity that does not depend on the coordinates in either configuration space or Fourier (momentum) space. The wave function of the state vector is obtained contracting it on the coordinate basis of either of these spaces. Hence, $$\psi(r)=\langle r|\psi\rangle$$ or $$\psi(p)=\langle p|\psi\rangle$$. Now $$\psi(r)$$ and $$\psi(p)$$ are related by a Fourier transform.

The assumption is that $$|r\rangle$$ or $$|p\rangle$$ represent complete orthogonal bases. Then one can always insert an identity given by $$I = \int |r\rangle \langle r| dr$$ between the state vector and the operator. It then gives $$\int \langle \psi_a|\hat{V}(t)|\psi_b\rangle dr = \int \langle \psi_a|r_1\rangle \langle r_1|\hat{V}(t)|r_2\rangle \langle r_2|\psi_b\rangle dr_1 dr_2 = \int \psi_a^*(r_1) V(r_1,r_2,t)\psi_b(r_2) dr_1 dr_2 .$$ If $$V$$ is diagonal (which means it is zero unless $$r_1=r_2=r$$), the expression then becomes $$\int \psi_a^*(r) V(r,t)\psi_b(r) dr .$$

• not sure I understand what you mean by “diagonal”. $V(r,t)$ depends only on $r$, not on $r_1,r_2$, and it can connect two different states $\psi_a$ and $\psi_b$. I don’t see how your answer encapsulates this. Dec 26, 2020 at 3:55
• You can think of $\langle i|\hat{A}|j\rangle$ as entries in a matrix $A_{ij}$, e.g. $|i\rangle=|\uparrow\rangle,|\downarrow\rangle$ and $\hat{A}=\hat{L}$, then you find $L_{ij}=\hbar/2\delta_{ij}$. In general we say $\hat{A}$ is orthogonal in a given basis ${|i\rangle}$, if $\langle i|\hat{A}|j\rangle\sim \delta_{ij}$. This idea can also be extended to uncountable basis such as ${|r\rangle}$. Dec 26, 2020 at 10:08

$$\int_{0}^{t}\left<\psi_{b}(r)|V(r,t)|\psi_{a}(r)\right>e^{i\omega_{0}t}dt$$ is really a shorthand for writing

$$\int_{0}^{t}dV\, \psi_b(r)^* V(r,t)\psi_{a}(r)e^{i\omega_{0}t}dt$$ with some integration over some volume (or maybe just $$rdr$$ or even $$r^2 dr$$ depending on the context). The notation avoids writing the spatial part of that integral.

I am unsure how to represent $$\psi_n(\phi) = \frac{1}{\sqrt{2 \pi}} e^{in\phi}$$ in the form $$\langle \psi_{n'}(r) |$$ and $$| \psi_n(r) \rangle$$ as required for (3).

I believe that $$\psi_n(\phi) = \frac{1}{\sqrt{2 \pi}} e^{in\phi}$$ is meant to denote just the angular part of the total wavefunction $$| \psi_n(r) \rangle$$. It's sufficient to know just the angular part to find out when (3) is non-zero.