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When I was reading the original paper about Runge-Gross theorem (Phys. Rev. Lett. 52, 997 (1984)), I saw $\hat{\psi} $ operator notation.

I've never seen these notations from my QM text, and I'm confused now.

In this paper, each operator consisting of the Hamiltonian $ \hat{H}(t) = \hat{T} + \hat{V}(t) + \hat{W} $ is assummed to have the form of

kinetic energy: $$ \hat{T} = \sum_{s} \int d^{3}r \hat{\psi}_{s}^{\dagger}(\vec{r}) \left( -\frac{1}{2} {\nabla}^{2} \right) \hat {\psi}_{s}(\vec{r}) $$

one-particle potential: $$ \hat{V} = \sum_{s} \int d^{3}r \hat{\psi}_{s}^{\dagger}(\vec{r}) v(\vec{r} \; t) \hat {\psi}_{s}(\vec{r}) $$

two-particle potential: $$ \hat{W} = \frac{1}{2}\sum_{s}\sum_{s'} \int d^{3}r \int d^{3}r' \hat{\psi}_{s}^{\dagger}(\vec{r}) \hat{\psi}_{s}^{\dagger}(\vec{r}') w(\vec{r'}, \vec{r}) \hat {\psi}_{s}(\vec{r}') \hat {\psi}_{s}(\vec{r}) $$

At first I thought $ \hat{\psi}_{s} \equiv \langle \mathbf{r}_{s}| $, but it seemed wrong when I plugged in above formula, and calculated $ \langle \psi|\hat{T} | \psi \rangle $.

Actually, I found a similar question posted earlier (What exactly is $\hat{\psi}^\dagger(x)$? An operator or a function?), but it doesn't give a clear answer for that.

I want to know textbooks or articles about the operator $\hat{\psi}_{s}$, and how to operate it manually when it applied to a physical state $ |\psi \rangle $.

Emilio Pisanty
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Patche
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1 Answers1

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You have encountered a formalism, that for better or worse (probably for worse) has come to be associated with the name second quantization. This formalism is primarily designed for situations where you can have one-particle wavefunctions $\psi(x)$, but also two-, three-, and even $n$-particle states with wavefunctions $\psi(x_1,x_2)$, $\psi(x_1,x_2,x_3)$ or $\psi(x_1,x_2,\ldots,x_n)$, and so on, and it enables you to treat the cases of arbitrary $n$ (and even superpositions of different $n$) within one unified setting.

(The name 'second quantization' is misleading and unfortunate. Nothing is getting "quantized again", and the appearance of hats on $\psi$s is purely cosmetic. The 'first-quantization' and 'second-quantization' formalisms are essentially equivalent and the former fits in unmodified as the special case $n=1$ of the latter.)

Similarly, finding a good text for second quantization that gets it just right can be rather tricky, and there are plenty of terrible expositions that mangle the subject, get way more technical than is really necessary, make rather misleading statements about what is 'quantized' and what isn't, use really unhelpful notation, and a number of other pitfalls. The Wikipedia page is reasonably passable, and there is this resource-recommendation question, to get you started.

Returning to $\hat \psi^\dagger(x)$, though: in short, this describes an operator that takes an $n$-particle position eigenstate $| x_1,\ldots,x_n\rangle$, and it transforms it into an $(n+1)$-particle position eigenstate by inserting an additional particle at position $x$, i.e., at the argument of $\hat\psi(x)$: $$ \hat \psi^\dagger(x)| x_1,\ldots,x_n\rangle = | x_1,\ldots,x_n,x\rangle. $$ Thus, $\hat\psi^\dagger(x)$ "creates" a particle at $x$. Similarly, and following from the above, $\hat \psi(x)$, "destroys" a particle at $x$, in some suitable sense.

You should be careful, however, with normalization, and with the bosonic or fermionic symmetry of the particles - the state $| x_1,\ldots,x_n\rangle$ is not equal to $| x_1\rangle \otimes \ldots \otimes |x_n\rangle$, but rather to its symmetrized or antisymmetrized version. This should make it clearer that the formula above is a huge simplification of what would otherwise be a mess of symmetrizers, Slater determinants, and whatnot.

Emilio Pisanty
  • 137,480