# Attempt to define a momentum space wavefunction for a superposition of of 1-particle states

Consider an arbitrary state in the Fock space constructed by superposing 1-particle states: $$|\psi\rangle=\mathbb{1}|\psi\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}|p\rangle\langle p|\psi\rangle$$ where I used the completeness relation for 1-particle states. Let me call $\langle p|\psi\rangle\equiv\psi(p)$. Therefore, $$|\psi\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}\psi(p)|p\rangle$$ I'm interested in calculating the object $\langle 0|\phi(x)|\psi\rangle$. Since, $\langle 0|\phi(x)|p\rangle=e^{ip\cdot x}$, I get, $$\langle 0|\phi(x)|\psi\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}\psi(p)\langle 0|\phi(x)|p\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}\psi(p)e^{ip\cdot x}$$

$\bullet$ How to invert this expression to express $\psi(p)$ (in terms of an integral over the LHS expression)?

$\bullet$ Can $\psi(p)$ be regarded as the momentum space wavefunction corresponding to the state $|\psi\rangle$?

• 1. In general, you can't "invert" integrals. 2. What is your requirement for something being able to be regarded as "the momentum space wavefunction", and why can you not check those properties yourself? – ACuriousMind Nov 26 '16 at 13:47
• @ACuriousMind- 2. $\psi(p)$ should be square integrable with Born interpretation. If $\psi(p)$ is not invertible how can I check that? Moreover, in the context of non-relativistic quantum mechanics, and with $\mathbb{1}=\int dp |p\rangle\langle p|$, $\psi(p)=\langle p|\psi\rangle$ is called momentum space wavefunction. So I wonder whether this analogy works here. – SRS Nov 26 '16 at 13:56

## 1 Answer

Let $$\langle 0|\phi(x)|\psi\rangle=\int d^3\textbf{p}\ f(p)\ e^{ip\cdot x}\tag{1}$$ where $$f(p)\equiv \frac{1}{(2\pi)^32E_p}\psi(p)$$

Multiply both sides of $(1)$ with $\mathrm e^{i\vec q\cdot\vec x}$, and integrate over $\mathrm d\vec x$: $$\int\mathrm d\vec x\ \mathrm e^{i\vec q\cdot \vec x}\langle 0|\phi(x)|\psi\rangle=\int d^3\textbf{p}\ f(p)\ \mathrm e^{iE_pt}\int\mathrm d\vec x\ e^{i(\vec q-\vec p)\vec x}$$

The integral of the exponential generates a delta function, and so $$\int\mathrm dx\ \mathrm e^{-i\vec q\cdot\vec x}\langle 0|\phi(x)|\psi\rangle=(2\pi)^3\int d^3\textbf{p}\ f(p)\mathrm e^{iE_pt}\delta(\vec q-\vec p)$$ so that $$\int\mathrm dx\ \mathrm e^{-i\vec q\cdot\vec x}\langle 0|\phi(x)|\psi\rangle=(2\pi)^3 f(q)\mathrm e^{iE_qt}$$

This means that $$\psi(p)=2E_p\int\mathrm dx\ \mathrm e^{ip x}\langle 0|\phi(x)|\psi\rangle$$

Your second question is meaningless unless you define what you mean by "momentum space wavefunction". You can for example define it by the expression above, in which case the answer is trivially yes: that expression is the momentum space wavefunction, by definition.

• In the comment above addressed at ACuriousMind I defined what I understand by a wavefunction i.e., a function which is square integrable and obeys Born interpretation of probability. – SRS Nov 26 '16 at 14:04
• @SRS the square integrability depends on the details of $|\psi\rangle$. For example, for a one-particle state $\psi(p)$ is not square integrable, but for wave packets in general it is square integrable. As for your second point, it is far from clear what the Born interpretation means in QFT. The question is still meaningless unless you properly define what you mean by Born interpretation of probability. – AccidentalFourierTransform Nov 26 '16 at 14:15
• Using $\langle \psi|\psi\rangle=1$, I could derive $\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}\psi^*(p)\psi(p)=1$. – SRS Nov 26 '16 at 14:28
• @SRS sure. $\langle\psi|\psi\rangle=1$ is equivalent to square integrability of $\psi(p)$. The issue is that many kets do not satisfy $\langle\psi|\psi\rangle=1$. For example, a one-particle state has $\langle p|p\rangle=\delta(0)=\infty$, and therefore $\psi(p)$ is not square integrable! – AccidentalFourierTransform Nov 26 '16 at 14:35
• @AccidentalFourierTransform "Multiply both sides of (1) with $e^{i\vec{q}\cdot \vec{x}}$, and integrate over $\mathrm{d}\vec{x}.$" I assume this was an accident? (I am sorry you must get that a lot.) – Brian Moths Nov 26 '16 at 16:14