# Wave functional

1. Aug 12, 2007

### jostpuur

Is it correct to think, that with a scalar complex Klein-Gordon field the wave function $$\Psi:\mathbb{R}^3\to\mathbb{C}$$ of one particle QM is replaced with an analogous wave functional $$\Psi:\mathbb{C}^{\mathbb{R}^3}\to\mathbb{C}$$? Most of the introduction to the QFT don't explain anything like this, but when I've thought about it myself, that seems to be correct.

If this was correct for the Klein-Gordon field, then the real problem is the Dirac's field. I don't understand what kind of wave functional it could have.

2. Aug 12, 2007

### dextercioby

The classical complex scalar field is a complex function defined on all the Minkowski space.

$$\varphi \in C^{\infty}\left(\mathbb{R}^{4},\mathbb{C}\right)$$

3. Aug 12, 2007

### jostpuur

But when we want a quantum mechanical field, we want to have complex amplitudes for all possible configurations of the classical field. That means, a wave functional $\Psi[\varphi]$.

4. Aug 12, 2007

### dextercioby

When quantized such a classical field like the KG field becomes an operator valued distribution having as a domain a subset of the bosonic/fermionic Fock space. See the first 2 volumes of Reed & Simon for details.

5. Aug 12, 2007

### jostpuur

Making the field and the conjugate field operators seems to be analogous the making position and momentum operators in the particle QM.

But when position and momentum are made operators, there is also the state which can be represented with a wave function, and we can have representations of the operators also.

I understand that in QFT we have operators for fields, but shouldn't we also have representations for the states, that means, shouldn't we have these wave functionals? And also actual representations for the operators?

6. Aug 13, 2007

### Demystifier

Yes, QFT can be formulated in that way as well. However, it is not usual in practice, because what one usually measures are not field configurations but positions/momenta/energies of particles.

7. Aug 13, 2007

### dextercioby

The fields are not observable, we're completely free to describe them using any possible representation for the uniparticle Hilbert space. Usually it's chosen $L^2 \left(\mathbb{R}^3, d^3 p\right)$, just like in ordinary QM. Anyways, the most important thing to QFT is to give valid predictions for the observables and the matematicals means to do it are not that relevant.

8. Aug 14, 2007

### Demystifier

The irony is that the only really observable things are particle positions, which however are not described by a hermitian operator, contradicting one of the cornerstone axioms of quantum theory.
So, is quantum field theory a genuine quantum theory?

9. Aug 15, 2007

### jostpuur

Similarly as a classical position of a point particle is only an expectation value of the position of a quantum mechanical particle, isn't the classical electromagnetic field only an expectation value of the quantum field theoretical electromagnetic field?

Last edited: Aug 15, 2007
10. Aug 17, 2007

### Demystifier

Formally yes, but it is not really consistent for fermionic fields.

11. Aug 17, 2007

### jostpuur

It is precisely the fermionic fields that make me feel like not understanding what's happening with quantum fields. Unfortunate stuff.

12. Aug 19, 2007

### samalkhaiat

13. Aug 19, 2007

### jostpuur

If I have a function $\Psi:\mathbb{R}^n\to\mathbb{C}$, and vector notation $x_i\in\mathbb{R}$ where $i\in\{1,2,\ldots, n\}$. Then the partial derivatives are given a notation

$$\partial_i\Psi$$ or $$\partial_{x_i}\Psi$$.

If I have a functional $\Psi:\mathbb{R}^{\mathbb{R}^3}\to\mathbb{C}$, and a vector notation $\phi_x = \phi(x) \in \mathbb{R}$ where $x\in\mathbb{R}^3$. Then I would analogously give the partial derivatives a notation

$$\partial_x\Psi$$ or $$\partial_{\phi_x}\Psi$$.

Is this the same thing that $$\frac{\delta}{\delta \phi(x)}\Psi$$ means?

Last edited: Aug 19, 2007
14. Aug 19, 2007

### samalkhaiat

15. Aug 19, 2007

### jostpuur

This is very interesting. It could be that this equation is precisely what I'm after, but I don't understand it yet.

16. Aug 19, 2007

### samalkhaiat

17. Aug 19, 2007

### reilly

See the first chapters of Zee's QFT book -- or most books on Solid State physics. Then you will understand that QFT is simply an alternate formulation of standard QM, based on creation and destruction operators -- let's hear it for harmonic oscillators -- such that it's easy to deal with systems in which the numbers of various particles is not fixed. And, if you want to see bread and butter QFT, then look at the field of Quantum Optics, Mandel and Wolf's book for example. To get a hold of QFT, you must study both theory AND practice -- neglect of one or the other will get you all messed up.
Regards,
Reilly Atkinson

18. Aug 20, 2007

### Demystifier

Jostpuur, for some standard references on functional Schrodinger equation for fermionic fields, see Refs. [16,17,18] in my
http://xxx.lanl.gov/abs/quant-ph/0302152