I've got several questions regarding the so called second quantization of the Schrödinger equation.

My professor introduced the field operators for the Schrödinger field by simply stating them as follows: $$ \hat\psi (\vec{r},\xi)=\sum\limits _i \psi_i(\vec{r},\xi) \hat a_i\\ \hat\psi^\dagger (\vec{r},\xi)=\sum\limits _i \psi_i^\star (\vec{r},\xi)\hat a^\dagger_i $$ Where $\psi_i(\vec{r},\xi)$ are the time independent one particle wave functions and $\hat a_i,\, \hat a^\dagger_i$ the corresponding creation and annihilation operators.

Is there a way to explain, why one does this? If I understood correctly what I've been taught so far, in QFT one must find some way to quantize the fields obeying the field equation in question. I do, however, not quite understand why in this particular case it is done like this.

Shouldn't the $\psi_i(\vec{r},\xi)$ be the time dependent one particle wave functions? Because I thought the field operators for a system in a box look like this: $$ \hat\psi (\vec{r},\xi)\sim\int\text{d}^3k\ \exp(i\omega_k t-i\vec k\vec x) \hat a_k $$

My professor then proceeded to prove the (anti)commutation relations between the field operators, postulating the corresponding relations between the fermionic or bosonic creation and annihilation operators: $$ \left[\hat\psi (\vec{r},\xi);\hat\psi^\dagger (\vec{r}',\xi')\right]_\pm= \left[\sum\limits_i\psi_i (\vec{r},\xi)\hat a_i;\sum\limits_j\psi^\star_j (\vec{r}',\xi')\hat a^\dagger_j\right]_\pm =\sum\limits_i\psi_i (\vec{r},\xi)\psi^\star_i (\vec{r}',\xi')\\ =\delta(\vec{r}-\vec{r}')\delta _{\xi,\xi'} $$ Here I do not understand the last step. Is that conclusion possible? And shouldn't or couldn't one postulate the commutation relations between the field operators and arrive at the relations for the creation and annihilation operators?


3 Answers 3


Is this not simply the closure relation $\sum_{i}\psi _{i}(\mathbf{r},\xi )\psi _{i}^{\ast }(\mathbf{r}\prime ,\xi \prime )=<\mathbf{r,\xi }|{\{}\sum_{i}|\psi _{i}><\psi _{i}|{\}}|% \mathbf{r}\prime ,\xi \prime >=<\mathbf{r,\xi }|\mathbf{r% }\prime ,\xi \prime >$?


I recommend you to have a look at the book "Advanced Quantum Mechanics" by F. Schwabl. In the chapter "1.5 Field Operators" he describes precisely what you are asking!


If I understood correctly what I've been taught so far, in QFT one must find some way to quantize the fields obeying the field equation in question.

This is correct. In this particular case you start with the Lagrangian that has Schrödinger's equation as its EOM. The following turns out to be the correct one:

$$\mathcal{L} = i\psi^*\partial_{t}\psi - \frac{1}{2M}(\partial_{j}\psi^*)(\partial_j \psi).$$

(the $j$'s are summed over). Now to quantize, we can proceed in the way you suggested, i.e impose CCR on the field $\psi$ and $\psi^*$ and then express them in terms of the creation and annihilation operators. Here it turns out that you can define the creation and annihilation operators to be the Fourier transform of the fields $\psi$ and $\psi^{\dagger}$, since you can show that the Fourier transforms obey the same commutation relations as the creation and annihilation operators need to, i.e

$$[\hat{\psi}(\mathbf{k}),\hat{\psi}^{\dagger}(\mathbf{k'})] = (2\pi)^3\delta(\mathbf{k}-\mathbf{k'}).$$

Though one thing I'm unsure why your professor did was summation instead of integration, since creation and annihilation operators usually depend on a continuous degree of freedom $k$ and thus need to be integrated over. That is, if I was to take that approach, I could postulate that $$\psi(\mathbf{x}) = \int \frac{d^3\mathbf{k}}{(2\pi)^3} e^{i\mathbf{k\cdot x}}a_{\mathbf{k}},$$

and from there show that $\psi$ and $\psi^{\dagger}$ obey the correct commutation relations if I impose the commutation relations for the $a$'s.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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