Need help regarding the derivation of a 2-particle wavefunction

[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.

 

[WWCY]

could someone assist? Many thanks

 

[TSny]

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

$| x,y \rangle = |x\rangle \otimes |y\rangle = \Big( \int dp' \phi_{p'}(x)|p'\rangle \Big) \otimes \Big( \int dq' \phi_{q'}(y)|q'\rangle \Big)$ 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]

Hi, thanks for the reply!

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? $|x,y\rangle = \frac{1}{\sqrt{2}} ( |x \rangle |y\rangle + |y \rangle |x\rangle )$

 

[TSny]

Hi, thanks for the reply!

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 $|x, y \rangle$.