# Second quantization of the Schrodinger fields

1. Oct 11, 2013

### AlbertEi

Hi,

I'm reading www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdf [Broken] and trying to understand the canonical quantization of the Schrodinger field. In particular, the Lagrangian:

\mathcal{L} = \frac{i}{2}\psi^* \partial_0 \psi - \frac{i}{2}\psi \partial_0 \psi^* + \frac{\left(\partial_i \psi^*\right) \left(\partial^i \psi\right)}{2m}

results in the Schrodinger equation for a free particle. Now, the canonical momentum is given by:

\Pi = \frac{\partial \mathcal{L}}{\partial \left(\partial_0 \psi\right)}= \frac{i}{2} \psi^*

and:

\Pi^* = \frac{\partial \mathcal{L}}{\partial \left(\partial_0 \psi^* \right)}= - \frac{i}{2} \psi

Subsequently, imposing the canonical commutation relation, I obtain:

[ \psi(\mathbf{x}_1,t),\Pi(\mathbf{x}_2,t)] = [ \psi(\mathbf{x}_1,t),\frac{i}{2}\psi^*(\mathbf{x}_2,t)] = i \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

and therefore:

[ \psi(\mathbf{x}_1,t),\psi^*(\mathbf{x}_2,t)] = 2 \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

However, it turns out I'm wrong by a factor of 2, because according to the link I provided (see equation (3.22)), the canonical commutation relation should be:

[ \psi(\mathbf{x}_1,t),\psi^*(\mathbf{x}_2,t)] = \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

Have I done something wrong in my calculations? Or am I supposed to renormalize the wave function such that the factor of 2 disappears? Or is there something else I am not understanding?

Any help would be much appreciated.

Last edited by a moderator: May 6, 2017
2. Oct 11, 2013

### stevendaryl

Staff Emeritus
Hmm. I checked with Hatfield (in "Quantum Field Theory of Point Particles and Strings") and he claims that for the Lagrangian

$\mathcal{L} = \frac{i}{2}(\psi^* \dot{\psi} - \dot{\psi^*}\psi) + ...$

the canonical momentum is $i \psi^*$, not $\frac{i}{2} \psi^*$

I don't know how he got that. However, I do know that you can use a different Lagrangian

$\mathcal{L'} = \mathcal{L} + \frac{i}{2} \frac{d}{dt}(\psi^* \psi) = i (\psi^* \dot{\psi}) + ...$

Using $\mathcal{L'}$ definitely leads to the canonical momentum $i \psi^*$. In Lagrangian theory, adding or subtracting a total derivative is supposed to make no difference (it has no bearing on the equations of motion). So $\mathcal{L'}$ is supposed to be an equivalent Lagrangian.

Last edited by a moderator: May 6, 2017
3. Oct 11, 2013

### AlbertEi

Hmmm.... that is very interesting. So you don't think my calculations are wrong, but that the author just adds a total divergence term to influence the expression for the canonical momentum. I never realized that this was allowed and makes me rethink the idea of canonical momentum, because basically this means that the canonical momentum is not a fundamental property of the universe. Or am I wrong?

In other words, different Lagrangian describing the same equations of motion can have different canonical momentums. How do we know what is the correct Lagrangian?

4. Oct 11, 2013

### dauto

The second term in the Lagrangian also contributes to the canonical momentum despite the appearances doubling its value as can be seen by doing an integration by parts of the Lagrangian.

Last edited: Oct 11, 2013
5. Oct 11, 2013

### AlbertEi

Ahhh very clever. Thanks!

6. Oct 11, 2013

### DrDu

Interesting. I would be careful with the complex operators as they are not hermitian whence I won't suppose them to fulfill canconical commutation relations. Maybe you could try to work with real operators and see what happens.

7. Oct 11, 2013

### DrDu

This has also been realized by others
http://users.physik.fu-berlin.de/~schotte/QUANT/schroeder.pdf

Maybe the book by Eugen Fick "Grundlagen der Quantenmechanik" contains a deeper discussion as I remember him to discuss the second quantization of the Schroedinger field in great detail.
As is also the case with first quantization, this is not a unique procedure.
I also wonder whether the factor 2 really makes a difference. After all it only rescales hbar whose value can't be inferred by second quantization but has to be put in by hand.

8. Oct 11, 2013

### AlbertEi

Unfortunately my German is not as good as I wish it was so I can't read the article. I think you are right that the factor of 2 is not important for the final (conceptual) result, but these kind of things really annoy me for some reason. And I'm happy I asked this question on this forum, because the answer by dauto has been very helpful and I think it might be important to understand these type of tricks.

Also, I don't really understand your first comment. I remember having seen a discussion that it was quite normal for operators to be non-Hermitian in QTF?