4
$\begingroup$

In the book quantum theory of many-particle systems by Fetter and Walecka, section 1.2, equation (2.4), the Hamiltonian writes:

$$ \hat{H} = \int d^3 x \hat{\psi}^\dagger(x) T(x) \hat{\psi}(x) $$ The comment below the equation says:

The quantities $\hat{\psi}$ and $\hat{\psi}^\dagger$ are not wave functions, however, but field operators; thus in second quantization the fields are the operators and the potential and kinetic energy are just complex coefficients.

I am a bit confused because on the one hand:

let $\hat{\psi} = \sum_k e^{ikx} c_k$, and $T(x)$ as an operator $-\hbar^2\nabla^2/2m$, one can smoothly get the ordinary second quantization form $\hat{H}=\sum_k\frac{\hbar^2k^2}{2m}c_k^\dagger c_k$.

On the other hand (assume bosons for simplicity): $$ [\hat{\psi}(x),\hat{H}] = [\hat{\psi}(x), \int d^3 z \hat{\psi}^\dagger(z) T(z) \hat{\psi}(z)] = \int d^3 z\left([\hat{\psi}(x), \hat{\psi}^\dagger(z) ] T(z) \hat{\psi}(z) + \hat{\psi}^\dagger(z)[\hat{\psi}(x), T(z)]\hat{\psi}(z) + \hat{\psi}^\dagger(z) T(z)[\hat{\psi}(x), \hat{\psi}(z)]\right) = T(x) \hat{\psi}(x) $$ which is obtained using $[\hat{\psi}(x), \hat{\psi}^\dagger(z)]=\delta(x-z)$, $[\hat{\psi}(x), \hat{\psi}(z)]=0$, and assume T(x) is just a c number.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

There are two different sorts of operators that you have to keep in mind. One type of operators regards the ones that act on the Hilbert space of the quantum field (usually called the Fock space). The other sort are the operators that act on complex functions, such as the differentiation, multiplication and the Laplacian.

What is meant there is that $T(x)$ is not an operator that acts in the Hilbert space of the quantum field. It is an operator that acts on complex functions (and gets extended to formally act on quantum fields). In the computation you show you don't have to treat $T(x)$ as a number, you just have to assume it acts as a differential operator in the functions. In what follows I will denote the application of $T(z)$ in $\psi(z)$ (as a differential operator) by $T(z)[\psi(z)]$. Notice in particular that $T(z)$ differentiates only operators that depend on the variable $z$ (it does not affect $\psi(x)$, for instance). The computation then reads

$$ [\hat{\psi}(x),\hat{H}] = [\hat{\psi}(x), \int d^3 z \hat{\psi}^\dagger(z) T(z) \hat{\psi}(z)] \\ = \int d^3 z\left([\hat{\psi}(x), \hat{\psi}^\dagger(z) ] T(z) \hat{\psi}(z) + \hat{\psi}^\dagger(z)[\hat{\psi}(x), T(z)]\hat{\psi}(z) + \hat{\psi}^\dagger(z) T(z)[\hat{\psi}(x), \hat{\psi}(z)]\right)\\ = \int d^3 z\left(\delta(x-z) T(z) \hat{\psi}(z) + \hat{\psi}^\dagger(z)\hat{\psi}(x)T(z)[\hat{\psi}(z)]- \hat{\psi}^\dagger(z)T(z)[\hat{\psi}(x)\hat{\psi}(z)] + \hat{\psi}^\dagger(z) T(z)[\hat{\psi}(x) \hat{\psi}(z)]-\hat{\psi}^\dagger(z)\hat{\psi}(x) T(z)[\hat{\psi}(z)]\right).$$

Now, the first term will give the desired result of $T(x)\hat{\psi}(x)$, but in the other terms you have a differential operator that differentiates with respect to $z$, so that the functions of $x$ are unaffected. That is, $T(z)[\hat{\psi}(x) \hat{\psi}(z)]=\hat{\psi}(x) T(z)[\hat{\psi}(z)]$. Thus, the remaining terms cancel.

$\endgroup$
6
  • $\begingroup$ In your derivation, I think you should consider the cases when $z$ equals to $x$, since $z$ run over all possible coordinates . Rigorously, you should not say $T(z)$ don't operate $\hat{\psi}(x)$.. $\endgroup$
    – modest
    Aug 9, 2021 at 2:38
  • $\begingroup$ $T(z)$ is a differential operator that does not act on the $x$ variable. For instance, if you were considering $\int dz\,\partial_z(f(x)g(z))$ the result would be simply $f(x)\int dz\,g'(z)$. This is exactly the same case here. $\endgroup$
    – Rick
    Aug 9, 2021 at 2:43
  • $\begingroup$ Moreover, neither $x$ or $z$ should be thought of a specific value before you integrate: it does not make sense do evaluate a differential operator at a point (e.g. $\frac{\partial}{\partial 2}$) $\endgroup$
    – Rick
    Aug 9, 2021 at 2:48
  • 1
    $\begingroup$ You are right, thanks. $\endgroup$
    – modest
    Aug 9, 2021 at 2:52
  • 1
    $\begingroup$ Yes. In order to conduct differentiation, you cannot have a function evaluated at a point. You must think of it as a function, and take the corresponding limit ($f'(2) = \lim_{\epsilon\rightarrow 0} \frac{f(2+\epsilon)-f(2)}{\epsilon}$, while $\partial_x [f(2)] = 0$, for instance). After finding the corresponding derivative of the functions, you should then integrate and consider "evaluating" the integrand in order to perform the sum (integral). $\endgroup$
    – Rick
    Aug 9, 2021 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.