All number of bosons in case of Bose condensation

[Petar Mali]

All number of bosons in case of Bose condensation is

$$N=N_0+\sum_{\vec{p}\neq 0}N_{\vec{p}}$$

I found in one book that $$N_0=\hat{b}^+_0\hat{b}_0$$

In that book author say that $$\hat{b}^+_0\hat{b}_0=N_0$$ and $$\hat{b}_0\hat{b}^+_0=N_0+1\approx N_0$$

so

$$[\hat{b}_0,\hat{b}^+_0]\approx 0$$

and from there

$$\hat{b}_0=\hat{b}^+_0=\sqrt{N_0}$$

Is this true? Little seems ridiculous to me because I do not know how he could get

$$\hat{b}_0\hat{b}^+_0=N_0+1\approx N_0$$


without using

$$[\hat{b}_0,\hat{b}^+_0]=1$$

So

$$1\approx 0$$

:)

Am I
right?

I have problem also with calling for example $$N_{\vec{p}}=\hat{b}^+_{\vec{p}}\hat{b}_{\vec{p}}$$ because $$N_{\vec{p}}$$ is eigen -value of this operator!

 

[xepma]

By
$$\hat{b}_0\hat{b}^+_0=N_0+1\approx N_0$$
it is implicitly ment that the approximation holds when you act on the wavefunction. So if $$|\Psi\rangle$$ represents the wavefunction, you normally have:$$\hat{b}^+_0\hat{b}_0|\Psi\rangle = N_0|\Psi\rangle$$
and using the commutation relation for the bosonic operators:
$$\hat{b}_0\hat{b}^+_0|\Psi\rangle = (N_0+1)|\Psi\rangle$$

but because $$N_0$$ is so big we can approximate these as equal.

So the authors are indeed sloppy, and strictly speaking you're right by saying that you should distinguish between operators and numbers. But in this case the eigenvalue is so big, we are allowed to replace the number operator by $$N_0$$. And for good reason, because it simplifies a lot!

 

[hiyok]

(continuing xepma) Put in other words, when two operators are approximately exchangible, they reduce to conventional numbers !