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This is my Hamiltonian. $\psi_{\alpha}$ is a bosonic field. $$H_{\alpha}=\int \mathrm{d} \mathbf{r} \psi_{\alpha}^{\dagger}(\mathbf{r})\left(-\frac{\nabla^{2}}{2 m}\right) \psi_{\alpha}(\mathbf{r})+\frac{V_{0}}{2} \int \mathrm{d} \mathbf{r} \psi_{\alpha}^{\dagger}(\mathbf{r}) \psi_{\alpha}^{\dagger}(\mathbf{r}) \psi_{\alpha}(\mathbf{r}) \psi_{\alpha}(\mathbf{r})$$

I'm interested in the potential term. I write it in bras and kets, and I get this:

$$\frac{V_{0}}{2}\left\langle\psi_{\alpha}\left|\left\langle\psi_{\alpha} | \psi_{\alpha}\right\rangle\right| \psi_{\alpha}\right\rangle$$

(Actually, I'm not 100% sure about this step, since I ignored the fact that there is only one integral, and for my step to work, I would have thought I would need a double integral).

Is $\left\langle\psi_{\alpha}\left|\left\langle\psi_{\alpha} | \psi_{\alpha}\right\rangle\right| \psi_{\alpha}\right\rangle = 1$? I know it would be normal quantum mechanics, but maybe that's not necessarily true in QFT with bosonic fields?

Any help would be much appreciated

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1 Answer 1

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It might be $1$ for a specific function, but it's not true in general, because the integrand is $|\psi_\alpha|^4$, so you would need to have that $\psi_\alpha$ is an unit vector both in $L^2$ and in $L^4$. Also: $$\left\langle\psi_{\alpha}\left|\left\langle\psi_{\alpha} | \psi_{\alpha}\right\rangle\right| \psi_{\alpha}\right\rangle=\int \mathrm{d}r \bar{\psi}_\alpha(r)\left(\int\mathrm{d}r' \bar{\psi}_\alpha(r')\psi_\alpha(r')\right)\psi_\alpha(r)$$ So it does not equal to your original integral.

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  • $\begingroup$ Thanks for answering! I see that I'm mistaken by equating the two expressions, but I still don't see why the bra-ket expression is not equal to 1. Surely the inner braket is equal to one, and then the outer one would collapse in, also making it one? $\endgroup$
    – user45757
    Jun 16, 2020 at 10:07
  • $\begingroup$ @user45757 Your braket expression is indeed $1$. I just pointed out why does it not equal to your original expression. $\endgroup$
    – Botond
    Jun 16, 2020 at 10:10
  • $\begingroup$ "It's not true in general, because the integrand is |ψα|4, so you would need to have that ψα is an unit vector both in L^2 and in L^4." It's this sentence which I don't understand $\endgroup$
    – user45757
    Jun 16, 2020 at 10:13
  • $\begingroup$ @user45757 the integrand in your "potential integral" equals to $|\psi_\alpha|^4$, so your integral is the 4th power of the $L^4$ norm of $\psi_\alpha$. It can happen that both the $L^2$ norm and the $L^4$ norm is one, for example in 1D, if $\psi_\alpha(x)=1$ on $[0,1]$ and zero otherwise, but usually the two norms are different. Is it better now? $\endgroup$
    – Botond
    Jun 16, 2020 at 10:19
  • $\begingroup$ @user45757 Comments on Physics SE support Mathjax, so it is preferred to use that rather than trying to type special characters. $\endgroup$ Jun 16, 2020 at 10:33

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