I am reading Modern Many-Particle Physics by Lipparini. In chapter 1.5 he talks about the matrix elements of the one-body operator: $$F_1 = \sum_{i=1}^{N}f(x_i)$$ He mentions that the matrix elements are given by: $$ \langle \phi | F_1|\psi \rangle=\int dx_1...dx_N \phi^*(x_1,...,x_N)\sum_{i=1}^{N}f(x_i) \psi(x_1,...,x_N)=N\int dx_1...dx_N \phi^*(x_1,...,x_N)f(x_1) \psi(x_1,...,x_N)$$ and that this is true for both fermions and bosons. My doubt is in the equality between the two integrals. I understand how they are equal in the case for bosons, but why the equality holds for fermions?

For example, if we have $N=2$ then $\psi(x_1,x_2)=\phi_1(x_1)\phi_2(x_2)-\phi_1(x_2)\phi_2(x_1)$. Then, we would have: $$ \sum_{i=1}^{N}f(x_i) =[f(x_1)+f(x_2)] \psi(x_1,x_2) =\big( [f(x_1)\phi_1(x_1)]\phi_2(x_2)-\phi_1(x_2)[f(x_1)\phi_2(x_1)] \big) +\big( \phi_1(x_1)[f(x_2)\phi_2(x_2)]-[f(x_2)\phi_1(x_2)]f(x_1)\phi_2(x_1) \big) $$ inside the first integral. But this is not equal to twice the first term (only applying $f(x_1)$.

Of course, am assuming I am doing something wrong, but what is it?

  • 1
    $\begingroup$ The original equation has two wave functions in the integral, while in you N=2 case you only include one wave function. What happened to the phi* term? $\endgroup$
    – ragnar
    Commented Feb 7, 2021 at 4:34

1 Answer 1


I think that you do not have to write the many-body wave function explicitly in terms of the single-particle wave functions (although you're of course allowed to do so). As already mentioned in the comments by ragnar, you did not calculate a matrix element.

Consider the following integral

$$\int \mathrm{d}x_1\,\mathrm{d}x_2 \, \phi^*(x_1,x_2)\, (f(x_1)+f(x_2)) \, \psi(x_1,x_2) \quad ,$$ where $\phi(x_1,x_2)$ and $\psi(x_1,x_2)$ are some wave functions of two identical fermions and hence anti-symmetrized. We can write the above integral as

$$ \int \mathrm{d}x_1\,\mathrm{d}x_2 \, \phi^*(x_1,x_2)\, f(x_1)\, \psi(x_1,x_2) +\int \mathrm{d}x_1\,\mathrm{d}x_2 \, \phi^*(x_1,x_2)\, f(x_2)\, \psi(x_1,x_2) $$

and due to the properties of the wave functions, this equals to $$2 \, \int \mathrm{d}x_1\,\mathrm{d}x_2 \, \phi^*(x_1,x_2)\, f(x_1)\, \psi(x_1,x_2) \quad , $$ which is exactly the special case $N=2$ of the equation you gave for $\langle \phi|F_1|\psi\rangle$. Note that the same equation holds for wave functions of bosons (i.e. symmetrized wave functions), as you correctly stated.


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