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For times sake, I will only write about the non-interacting part of the Hamiltonian, $$H_0=\sum_{j=1}\left(-\frac{\hbar^2}{2m}\frac{\partial}{\partial x_j^2}+U(x_j)\right)$$ where $U(x_j)$ is some scalar potential.

In second quantised language the wave function is given by $$|\phi_\alpha\rangle=\int dx_1\cdots dx_N\,\phi_\alpha(x_1,\cdots,x_N)\psi^T(x_1)\cdots\psi^T(x_N)|0\rangle.$$

The Hamiltonian: $$H_{0,s}=\int dx\,\psi^T(x)H_0\psi(x).$$

SHOW $$H_{0,s}|\phi_\alpha\rangle=\int dx_1\cdots dx_N\,(H_0\phi_\alpha(x_1,\cdots,x_N))\psi^T(x_1)\cdots\psi^T(x_N)|0\rangle.$$

I then set the basis vectors to be the wavefunction $\psi_\alpha$ for conveinence.

So, I started by simply writing out: $$\int dx\,\psi^T(x)H_0\psi(x)\int dx_1\cdots dx_N\,\phi_\alpha(x_1,\cdots,x_N)\psi^T(x_1)\cdots\psi^T(x_N)|0\rangle$$ and using the definition of the position dependent density operator $\rho(x)=\psi^T(x)\psi(x)$. Since the basis vectors are the wave function, we can commute $H_o$ around and get $$\int dx\,\psi^T(x)\psi(x)H_0=\int dx\,\rho(x)H_0=H_0$$ since we are in 1D in a 1 particle state, the volume integral over the density will just 1. Then one can just use the same thing of the basis vectors to get $H_0$ into place in the final expression.

This is a homework question, and I don't feel satisfied with this as an answer; is it correct, or completely wrong? I'm not sure about the justification for the commutation of $H_0$, it feels too flimsy and my grasp of the linear algebra required here is somewhat lacking.

Also where here can we consider the difference between bosons and fermions? I have derived the commutation law $[\psi(x),\psi^T(y)]=\delta(x-y)$, but i don't know where to use it.

Effectively, I have some idea of whats going on, but not a complete one - and I think I've been sat with it for too long to see where I'm going wrong

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  • $\begingroup$ You can't commute field operators past $\hat{H}_0$, because $H_0$ has spatial derivatives, and the field operators depend on position. $\endgroup$
    – march
    Commented Apr 16, 2019 at 15:57
  • $\begingroup$ I also think you definition of $H_{0,s}$ is wrong. It shouldn't involve the sum over all $j$ particles. The correct definition would be something like $H_{0,s} = \int d^3x\hat{\psi}^{\dagger}(\vec{x})\left({T}(\vec{x}) + {U}(\vec{x})\right)\hat{\psi}(\vec{x})$, without the sum. $\endgroup$
    – march
    Commented Apr 16, 2019 at 16:04
  • $\begingroup$ @march $H_{o,s}$ is 2nd quantised language is more or less as you said, in the third equation down! and what do you mean "cant commute pas $H_o$"? does it literally refer to the order in the equation? $\endgroup$
    – Learn4life
    Commented Apr 16, 2019 at 16:23
  • $\begingroup$ When I mean "commute past", I mean to take advantage of the commutator to exchange the order of operators. So, for instance, $\psi(\vec{x})\psi^{\dagger}(\vec{x_1}) = [\psi(\vec{x}),\psi^{\dagger}(\vec{x_1})] + \psi^{\dagger}(\vec{x_1})\psi(\vec{x}) = \delta(\vec{x}-\vec{x}_1) + \psi^{\dagger}(\vec{x_1})\psi(\vec{x})$. However, frankly, I'm a little confused about the intended final result. I'm not really sure how you're supposed to move the derivatives in $H_0$ to acting on $\phi_{\alpha}$ when they're originally acting on $\psi(\vec{x})$. $\endgroup$
    – march
    Commented Apr 16, 2019 at 16:26
  • $\begingroup$ me too, thats i guess my question..is it to do with choosing a basis? if we choose the ψs as a basis, then doesnt this fix the position, we can commute stuff around? $\endgroup$
    – Learn4life
    Commented Apr 16, 2019 at 16:31

1 Answer 1

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Let us define: $$ \boxed{\hat{H}_0(x) = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(x)} $$ and $$ \boxed{\hat{H}_{0,s} = \int dx\,\Psi^T(x)\hat{H}_0(x)\Psi(x)} $$ When there are $N$ decoupled particles the total Hamiltonian takes the form: $$ \boxed{\hat{H}_0 = \sum_{i=1}^N\hat{H}_0(x_i)} $$ Notice that it is the single particle Hamiltonian, $\hat{H}_0(x)$, that appears in the definition of $\hat{H}_{0,s}$ and not $\hat{H}_0$ as the OP writes. Indeed, (as has been pointed out in the comments) since there is an integral over $x$ in $\hat{H}_{0,s}$ it does not make sense to insert $\hat{H}_0$ in $\hat{H}_{0,s}$.

We then define the multiparticle state, $|\phi_a\rangle$, with corresponding wavefunction, $\phi_a(x_1,\dots,x_N)$, by: $$ \boxed{|\phi_a\rangle=\int dx_1\dots dx_N\phi_a(x_1,\dots,x_N)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle} $$

Then, using that $[\Psi(x),\Psi^T(y)]=\delta(x-y)$, \begin{equation} \begin{aligned} \hat{H}_{0,s}&|\phi_a\rangle = \int dx\,\Psi^T(x)\hat{H}_0(x)\Psi(x)\int dx_1\dots dx_N\phi_a(x_1,\dots,x_N)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &=\int dx\,\Psi^T(x)\hat{H}_0(x)\int dx_1\dots dx_N\phi_a(x_1,\dots,x_N)\sum_{i=1}^N\delta(x-x_i)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ &=\int dx\,\Psi^T(x)\hat{H}_0(x)\sum_{i=1}^N\int dx_1\dots dx_i\!\!\!/\dots dx_N\phi_a(x_1,\dots,x,\dots,x_N)\Psi(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ &=\int dx\,\Psi^T(x)\sum_{i=1}^N\int dx_1\dots dx_i\!\!\!/\dots dx_N\big(\hat{H}_0(x)\phi_a(x_1,\dots,x,\dots,x_N)\big)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ &=\int dx_1\dots dx_N\Big(\sum_{i=1}^N\hat{H}_0(x_i)\phi_a(x_1,\dots,x_N)\Big)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &=\int dx_1\dots dx_N\Big(\hat{H}_0\phi_a(x_1,\dots,x_N)\Big)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ \end{aligned} \end{equation} which is what was meant to be shown. In the second to last step one changes variables of integration and in the last step we used the definition of $\hat{H}_0$ given above.


By request I'm adding more detail:

An important step in the above derivation is going from the first to the second equality. Note primarily that $\hat{H}_0(x)$ acts on $\Psi(x)$, which is the only quantity to its right that depends on $x$. Isolating the relevant terms then, and using only the commutator, $[\Psi(x),\Psi^T(y)]=\delta(x-y)$, \begin{equation} \begin{aligned} \big(\hat{H}_0(x)\Psi(x)\big)\Psi^T(x_1)&=\Psi^T(x_1)\hat{H}_0(x)\Psi(x)+\hat{H}_0(x)\delta(x-x_1)\\ &=\Psi^T(x_1)\hat{H}_0(x)\Psi(x)+\hat{H}_0(x)\delta(x-x_1)\\ \big(\hat{H}_0(x)\Psi(x)\big)\Psi^T(x_1)\Psi^T(x_2)&=\big(\Psi^T(x_1)\hat{H}_0(x)\Psi(x)+\hat{H}_0(x)\delta(x-x_1)\big)\Psi^T(x_2)\\ &=\Psi^T(x_1)\big(\hat{H}_0(x)\Psi^T(x_2)\Psi(x)+\hat{H}_0(x)\delta(x-x_2)\big)+\Psi^T(x_2)\hat{H}_0(x)\delta(x-x_1)\\ &=\Psi^T(x_1)\Psi^T(x_2)\hat{H}_0(x)\Psi(x)+\Psi^T(x_1)\hat{H}_0(x)\delta(x-x_2)+\Psi^T(x_2)\hat{H}_0(x)\delta(x-x_1)\\ &\vdots \end{aligned} \end{equation} Now $\Psi(y)$ is an annihilation operator, so that $\Psi(y)|0\rangle=0$. The above two equations when acting on the vacuum then read: \begin{equation} \begin{aligned} \big(\hat{H}_0(x)\Psi(x)\big)\Psi^T(x_1)|0\rangle &=\Big(\hat{H}_0(x)\delta(x-x_1)\Big)|0\rangle\\ \big(\hat{H}_0(x)\Psi(x)\big)\Psi^T(x_1)\Psi^T(x_2)|0\rangle &=\Big(\Psi^T(x_1)\hat{H}_0(x)\delta(x-x_2)+\Psi^T(x_2)\hat{H}_0(x)\delta(x-x_1)\Big)|0\rangle\\ \end{aligned} \end{equation} so the general result reads, \begin{equation} \big(\hat{H}_0(x)\Psi(x)\big)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle= \sum_{i=1}^N\Big(\hat{H}_0(x)\delta(x-x_i)\Big)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle \end{equation} Notice here that the combination $\hat{H}_0(x)\delta(x-x_i)$ can pass through the quantity $\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)$ without obstruction: they commute. Including now the multi-particle wavefunction contribution, \begin{equation} \begin{aligned} &\big(\hat{H}_0(x)\Psi(x)\big)\phi_a(x_1,\dots,x_N)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &= \sum_{i=1}^N\Big(\hat{H}_0(x)\delta(x-x_i)\Big)\phi_a(x_1,\dots,x_N)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ &= \sum_{i=1}^N\Big(\hat{H}_0(x)\delta(x-x_i)\phi_a(x_1,\dots,x_N)\Big)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ &= \sum_{i=1}^N\Big(\hat{H}_0(x)\delta(x-x_i)\phi_a(x_1,\dots,x,\dots,x_N)\Big)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ \end{aligned} \end{equation} where in the last equality we used the delta function to replace the $x_i$ argument of the wavefunction by $x$.

Let me mention in passing that one can also write $\hat{H}_0(x)\delta(x-x_i)=\hat{H}_0(x_i)\delta(x-x_i)$ and then integrate by parts twice in $x_i$ after having included the $x_i$ integral, and this would then also make manifest that $\hat{H}_0(x_i)$ acts on the wavefunction -- this leads to the same result. To show that all such approaches lead to the same result it suffices to realise that all of these are equivalent: $$ \int dx_i\Big(\frac{d^2}{dx^2}\delta(x-x_i)\phi(x_i)\Big)=\frac{d^2}{dx^2}\int dx_i\Big(\delta(x-x_i)\phi(x_i)\Big)=\frac{d^2}{dx^2}\phi(x) $$ $$ \int dx_i\Big(\frac{d^2}{dx^2}\delta(x-x_i)\phi(x_i)\Big)=\int dx_i\Big(\frac{d^2}{dx^2_i}\delta(x-x_i)\phi(x_i)\Big)=\int dx_i\Big(\delta(x-x_i)\frac{d^2}{dx^2_i}\phi(x_i)\Big)=\frac{d^2}{dx^2}\phi(x) $$ $$ \int dx_i\Big(\frac{d^2}{dx^2}\delta(x-x_i)\phi(x_i)\Big)=\int dx_i\frac{d^2}{dx^2}\Big(\delta(x-x_i)\phi(x)\Big)=\frac{d^2}{dx^2}\int dx_i\delta(x-x_i)\phi(x)=\frac{d^2}{dx^2}\phi(x) $$ In the last equation there is no inconsistency (as one can check by using the product rule for differentiation and then evaluating the $x_i$ integral taking into account that the integral commutes with the derivatives - all terms vanish except for the term yielding $\frac{d^2}{dx^2}\phi(x)$.) Since all of these lead to the same result after integrating out $x_i$, we can effectively replace $(\frac{d^2}{dx^2}\delta(x-x_i))\phi(x_i)$ by $\frac{d^2}{dx^2}(\delta(x-x_i)\phi(x))$ which is what we did above.

Next include also the measure contributions, \begin{equation} \begin{aligned} &\big(\hat{H}_0(x)\Psi(x)\big)\int dx_1\dots dx_N\,\phi_a(x_1,\dots,x_N)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &= \sum_{i=1}^N\int dx_1\dots dx_i\!\!\!/\,\,\,\dots dx_N\Big(\int dx_i\hat{H}_0(x)\delta(x-x_i)\phi_a(x_1,\dots,x,\dots,x_N)\Big)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ \end{aligned} \end{equation} and integrate out $x_i$ in the $i^{\rm th}$ term in the sum (using the delta function), \begin{equation} \begin{aligned} &\big(\hat{H}_0(x)\Psi(x)\big)\int dx_1\dots dx_N\,\phi_a(x_1,\dots,x_N)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &= \sum_{i=1}^N\int dx_1\dots dx_i\!\!\!/\,\,\,\dots dx_N\Big(\hat{H}_0(x)\phi_a(x_1,\dots,x,\dots,x_N)\Big)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ \end{aligned} \end{equation} Including now the factor $\Psi^T(x)$ and integrating over $x$ yields $\hat{H}_{0,s}|\phi_a\rangle$, \begin{equation} \begin{aligned} \hat{H}_{0,s}|\phi_a\rangle=&\big(\int dx \,\Psi^T(x)\hat{H}_0(x)\Psi(x)\big)\int dx_1\dots dx_N\,\phi_a(x_1,\dots,x_N)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &= \sum_{i=1}^N\int dx_1\dots dx_i\!\!\!/\,\,\,\dots dx_N\int dx\,\Psi^T(x)\Big(\hat{H}_0(x)\phi_a(x_1,\dots,x,\dots,x_N)\Big)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ \end{aligned} \end{equation} Next redefine the integration variable, $x\rightarrow x_i$, use that the $\Psi^T(x_i)$'s commute among themselves, and that the combination $\hat{H}_0(x)\phi_a(x_1,\dots,x,\dots,x_N)$ is just a function, so that $\Psi^T(x)$ can pass through it without obstruction: \begin{equation} \begin{aligned} &\hat{H}_{0,s}|\phi_a\rangle= \sum_{i=1}^N\int dx_1\dots dx_i\!\!\!/\,\,\,\dots dx_N\int dx_i\,\Big(\hat{H}_0(x_i)\phi_a(x_1,\dots,x_i,\dots,x_N)\Big)\Psi^T(x_i)\Psi^T(x_1)\dots\Psi^T(x_i\!\!\!/)\dots \Psi^T(x_N)|0\rangle\\ &= \sum_{i=1}^N\int dx_1\dots dx_N\,\Big(\hat{H}_0(x_i)\phi_a(x_1,\dots,x_N)\Big)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &= \int dx_1\dots dx_N\,\Big(\sum_{i=1}^N\hat{H}_0(x_i)\phi_a(x_1,\dots,x_N)\Big)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ &= \int dx_1\dots dx_N\,\Big(\hat{H}_0\phi_a(x_1,\dots,x_N)\Big)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle\\ \end{aligned} \end{equation} where we used the definition of $\hat{H}_0$ given above. The resulting equation, $$ \boxed{\hat{H}_{0,s}|\phi_a\rangle= \int dx_1\dots dx_N\,\Big(\hat{H}_0\phi_a(x_1,\dots,x_N)\Big)\Psi^T(x_1)\dots \Psi^T(x_N)|0\rangle} $$ is the desired result.

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  • $\begingroup$ The issue that I'm having with this derivation is that in the definition of $\hat{H}_{0,s}$, $H_0$ (with its implicit derivatives) acts only on the $\psi(\vec{x})$ to its right and not on whatever ends up to the right of $\psi(\vec{x})$ when you act on the state $|\phi\rangle$. Where in your derivation is the derivative inside $H_0$ allowed to act on $\phi$? Is there some sort of implicit integration by parts or something? $\endgroup$
    – march
    Commented Apr 17, 2019 at 4:29
  • $\begingroup$ I will edit to make this point clear in the answer (there's no integration by parts). $\endgroup$ Commented Apr 17, 2019 at 7:48
  • $\begingroup$ @march I included all of the intermediate steps so as to make clear the reasoning at each step $\endgroup$ Commented Apr 17, 2019 at 8:45
  • $\begingroup$ @march Incidentally, the notation of the OP is inconsistent as you correctly pointed out in your second comment following the OP's question. There are a few ways of correcting this and I chose one such way, by redefining what one means by $\hat{H}_0(x)$ rather than redefining $\hat{H}_{0,s}$. $\endgroup$ Commented Apr 17, 2019 at 8:56
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    $\begingroup$ That's a lot of extra TeX you had to write. I appreciate the explanation. $\endgroup$
    – march
    Commented Apr 17, 2019 at 16:54

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