# Need help regarding the derivation of a 2-particle wavefunction

• WWCY
In summary, the factor of ##1/\sqrt{2!}## is needed to prevent double counting that results from unrestricted sums. This is because when you perform a Fourier transform on a state that is already symmetric, the result will not have the factor of ##\frac{1}{\sqrt{2}}##.
WWCY
Homework Statement
From Blundell's QFT
Relevant Equations
Fourier transforms
I have an issue trying to understand the derivation of equation 3.40 (screenshot attached) of Blundell's QFT book. Here's my attempt.

##| x,y \rangle = |x\rangle \otimes |y\rangle = \Big( \int dp' \phi_{p'}(x)|p'\rangle \Big) \otimes \Big( \int dq' \phi_{q'}(y)|q'\rangle \Big)## which gives ##\int dp' \int dq' \phi_{p'}(x) \phi_{q'}(y) |p',q'\rangle##.

I can't seem to understand the argument by Blundell that "the factor of ##1/\sqrt{2!}## is needed to prevent double counting that results from unrestricted sums". Why does the factor of ##1/\sqrt{2!}## not "pop out" on its own? Aren't Fourier transforms supposed to preserve normalisation?Cheers.

could someone assist? Many thanks

Since no one has responded yet, I will try to help a little. I am not very familiar with this topic.

WWCY said:
##| x,y \rangle = |x\rangle \otimes |y\rangle = \Big( \int dp' \phi_{p'}(x)|p'\rangle \Big) \otimes \Big( \int dq' \phi_{q'}(y)|q'\rangle \Big)## which gives ##\int dp' \int dq' \phi_{p'}(x) \phi_{q'}(y) |p',q'\rangle##.

You assumed here that ##|x,y \rangle = |x\rangle \otimes |y\rangle ## and ##|q,p \rangle = |p \rangle \otimes |q\rangle ##. But I don't think that's right. ##| x,y \rangle ## and ##| q,p \rangle## are two-particle states that are already symmetrized or antisymmetrized for bosons or fermions.

For example, the definition of ##| p,q \rangle## is ##| p,q \rangle = \hat a_p ^\dagger \hat a_q^ \dagger | 0 \rangle##. Using the commutation relations for the operators, you see that ##| p,q \rangle = \pm | q, p \rangle##, (+ for bosons, - for fermions). But, ##|p \rangle \otimes |q\rangle\neq \pm |q \rangle \otimes |p\rangle##.

I can't seem to understand the argument by Blundell that "the factor of ##1/\sqrt{2!}## is needed to prevent double counting that results from unrestricted sums". Why does the factor of ##1/\sqrt{2!}## not "pop out" on its own? Aren't Fourier transforms supposed to preserve normalisation?
Here, I don't think I can help much. As I see it, the factor of ##1/\sqrt{2!}## is part of the definition of ##|x,y \rangle##. I'm not sure of the interpretation of Blundell's remark about preventing double counting. But, with the factor of ##1/\sqrt{2!}##, you can see that you get the properly normalized result in Blundell's (3.40).

WWCY

I completely forgot about the part regarding anti-symmetry/symmetric states.

However, would it make any sense if I performed the Fourier transforms on this (symmetric) state? ##|x,y\rangle = \frac{1}{\sqrt{2}} ( |x \rangle |y\rangle + |y \rangle |x\rangle )##

WWCY said:

I completely forgot about the part regarding anti-symmetry/symmetric states.

However, would it make any sense if I performed the Fourier transforms on this (symmetric) state? ##|x,y\rangle = \frac{1}{\sqrt{2}} ( |x \rangle |y\rangle + |y \rangle |x\rangle )##

If you take ##|p, q \rangle## as corresponding to ##\frac{1}{\sqrt{2}}( |p \rangle |q \rangle + |q \rangle ||p \rangle)## and ##|x, y \rangle## as corresponding to ##\frac{1}{\sqrt{2}} (|x \rangle |y \rangle + |y \rangle |x \rangle )##, then ## \langle y, x | p, q \rangle## will not yield the factor of ##\frac{1}{\sqrt{2}}## in Blundell's (3.40). If you want to get that factor, it appears that you would have to take ##|x,y\rangle## as corresponding to ## \frac{1}{2} ( |x \rangle |y\rangle + |y \rangle |x\rangle )## which has an overall factor of ##\frac{1}{2}## instead of ##\frac{1}{\sqrt{2}}##. This corresponds to the extra factor of ##\frac{1}{\sqrt{2!}}## that Blundell includes in his definition of the change of basis from the basis ##|p, q \rangle ## to the basis ##|x, y \rangle##.

## 1. What is a 2-particle wavefunction?

A 2-particle wavefunction is a mathematical representation of the quantum state of a system consisting of two particles. It describes the probability amplitude of finding the two particles at different positions and times, and can be used to calculate various properties of the system.

## 2. How is the 2-particle wavefunction derived?

The 2-particle wavefunction is derived using the Schrödinger equation, which is a fundamental equation in quantum mechanics. It involves solving a set of differential equations to determine the wavefunction, which is a complex-valued function of the positions and times of the two particles.

## 3. What are the assumptions made in deriving the 2-particle wavefunction?

In deriving the 2-particle wavefunction, it is assumed that the particles are non-interacting and that the potential energy between them is constant. It is also assumed that the particles are distinguishable, meaning that they can be identified and tracked separately.

## 4. Can the 2-particle wavefunction be used for any type of particles?

Yes, the 2-particle wavefunction can be used for any type of particles, as long as they are non-interacting and the assumptions mentioned above hold. This includes particles with different masses, charges, and spin states.

## 5. What are some applications of the 2-particle wavefunction?

The 2-particle wavefunction is used in various fields of physics, such as quantum chemistry, nuclear physics, and solid state physics. It is also essential in understanding phenomena like entanglement and quantum teleportation, and plays a crucial role in the development of quantum technologies.

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