According to what I have read, the second quantisation originally came from the effort to quantise the many body wave function in the Schrodinger equation.

We could write down the commutation relation of $\psi$ and $\psi^{\dagger}$ by first applying

$$|\psi(t)\rangle = \sum_n \psi_n(t) |n\rangle $$

to the Schrodinger equation

$$H|\psi(t)\rangle=i\hbar \partial_t |\psi(t)\rangle$$

$$ \langle n|H| \psi(t) \rangle = \sum_m \psi_m(t) \langle n|H|m \rangle = i\hbar \dot \psi_n(t) $$

Then consider the Hamiltonian as a functional of $\psi$ and $\psi^{*}$, that is

$$H(\psi,\psi^*)=\langle H\rangle = \langle\psi|H|\psi\rangle = \sum_{m,n} \langle\psi| m \rangle \langle m|H|n \rangle \langle n|\psi \rangle = \sum_{m,n} \psi_m^* \psi_n \langle m|H|n \rangle$$

Hence, by doing partial differential on $H$ with respect to $\psi$ and $\psi^*$, we get

$$\dot{\psi_m} = \frac{\partial H}{\partial (i\hbar \psi_m^*)}$$

$$i\hbar\dot{\psi_m^*} = -\frac{\partial H}{\partial \psi_m}$$

This has the same form as the Hamilton equations. Therefore, regarding $\psi$ as an operator, we can identify $\hat\psi_n \rightarrow \hat q_n$ and $i\hbar \hat\psi_n^\dagger \rightarrow \hat p_n$.

The commutation relation between $\hat\psi$ and its conjugate is then

$$[\hat\psi,\hat\psi^\dagger] = 1$$

However, I still cannot really see what the operator $\hat\psi$ actually does.

I have seen in many textbooks and lecture notes that $\hat\psi$ is an operator that annihilates a particle from the many-body wave function. Why is that?

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    $\begingroup$ It would help to add the hat $\^$ sign to your operators, so that we can easily distinguish which $\psi$ is an operator, which $\psi$ is a label for a bra or a ket and which one is just a complex number. $\endgroup$ – wcc Jul 18 '18 at 15:08
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    $\begingroup$ An example: If $\hat{\psi}_k (r)$ is an annihilation operator at position $r$ associated with plane-wave with wavevector $k$, then $\langle 0|\hat{\psi}_k (r)|k\rangle = e^{ikr}$ , where $|0\rangle$ is the vacuum and $|k\rangle$ is a momentum eigenstate with wavevector $k$. $\endgroup$ – wcc Jul 18 '18 at 15:11
  • $\begingroup$ @IamAStudent Thank you for the suggestion, I have corrected those operators with the hat. In you example, I still quite don't get how you get the expression $\langle 0|\hat{\psi}_k (r)|k\rangle = e^{ikr}$ if $\hat{\psi}_k (r)$ is defined by what you said. However, if $\hat{\psi}_k (r)$ is just an annihilation operator that annihilates the particle "localised" in the position space at $r$, I can see that $\langle 0|\hat{\psi}_k (r)$ is basically $\langle r|$ and the expression above is just a plane wave solution in the position basis. $\endgroup$ – Akemi Homura Jul 18 '18 at 16:02
  • $\begingroup$ @IamAStudent By the way, would you please tell me why the action of $\hat{\psi}(r)$ is to annihilate the particle localised at $r$? $\endgroup$ – Akemi Homura Jul 18 '18 at 16:03

Second quantization is building a field of operators out of infinitely many first quantized harmonic oscillators, each placed at a position. A field of quantum harmonic oscillators...

Back in the first quantization, you have annihilation and creation operators acting on the momentum space. Let $| 0 \rangle$ be the vacuum state, so any single-particle momentum eigenstate is given by the following: $$ \tag{1} | p \rangle = \hat{a}^\dagger (p) \, | 0 \rangle $$ where $\hat{a}^\dagger (p)$ creates a mode with momentum $p$. This is the good ol' first quantization.

On the other hand, we can write the position eigenstate, which is the Fourier transform of momentum eigenstate, in terms of (1), as follows: \begin{align} \tag{2} | x \rangle &= \int dp \; \psi^\dagger (x) \;| p \rangle \\ \end{align} where $\psi^\dagger (x) =\langle p|x \rangle$ is the wave function (I omit normalization factors). Now let's substitude (1) here. \begin{align} \tag{2} | x \rangle &= \int dp \; \psi^\dagger (x) \;\hat{a}^\dagger (p) \, | 0 \rangle \\ \end{align} Now, the expression, $\psi^\dagger (x) \;\hat{a}^\dagger (p)$, is like a term in a weighted sum, where the sum corresponds to the integral over momentum space.

Intuitively, we can think this as each particle at position, $x$, is created as an oscillation packet with all momentum values superposed with the weight, $\psi^\dagger(x)$. So, considering all the positions, we have a field of superposed oscillators with all possible momentum values. We can write this field explicitly.

So, the field operator, $\hat{\Psi}^\dagger (x)$, is a creation operator at position $x$ such that, \begin{align} \tag{3} | x \rangle &= \hat{\Psi}^\dagger (x) \, | 0 \rangle \\ \end{align} where \begin{align} \tag{4} \hat{\Psi}^\dagger (x) = \int dp \; \psi^\dagger (x) \;\hat{a}^\dagger (p) \end{align} So, (3) tells that a particle at position $x$ is the creation of the superposition of infinitely many oscillators on the vacuum state, and the field operator is that operation for every $x$. You can see that the field operator, $\hat{\Psi}^\dagger (x)$, is a particle creation operator because it is basically the Fourier transform of the ladder creation operator, $\hat{a}^\dagger (p)$.

This is the intuitive construction of canonical second quantization. Now, from here, one can check the commutation relations of fields and their conjugate momenta, using the commutation relations of the ladder operators. Other details regarding the rigorous mathematical and physical correspondence can be achieved one by one.


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