Klein-Gordon Field QM: Replacing Wave Function with Wave Functional?

[jostpuur]

Is it correct to think, that with a scalar complex Klein-Gordon field the wave function [tex]\Psi:\mathbb{R}^3\to\mathbb{C}[/tex] of one particle QM is replaced with an analogous wave functional [tex]\Psi:\mathbb{C}^{\mathbb{R}^3}\to\mathbb{C}[/tex]? 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.

 

[dextercioby]

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

[tex] \varphi \in C^{\infty}\left(\mathbb{R}^{4},\mathbb{C}\right) [/tex]

 

[jostpuur]

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

[tex] \varphi \in C^{\infty}\left(\mathbb{R}^{4},\mathbb{C}\right) [/tex]

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 [itex]\Psi[\varphi][/itex].

 

[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.

 

[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?

 

[Demystifier]

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 [itex]\Psi[\varphi][/itex].

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.

 

[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 [itex] L^2 \left(\mathbb{R}^3, d^3 p\right) [/itex], 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.

 

[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?
For a possible answer see
https://www.physicsforums.com/showthread.php?t=171642

 

[jostpuur]

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 [itex] L^2 \left(\mathbb{R}^3, d^3 p\right) [/itex], 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.

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?

 

[Demystifier]

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?

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

 

[jostpuur]

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

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

 

[samalkhaiat]

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.

If you know the defining relationship of a functional derivative with respect to Grassmann function, then you can transform the Schrodinger equation

[tex]i \frac{\partial}{\partial t} |\Psi \rangle = H |\Psi \rangle [/tex]

into the one suitable for spinor field theory;

[tex]i \frac{\partial}{\partial t} \Psi [ \psi ,t] = \int dx^{3} \left[ \frac{\delta}{\delta \psi (\vec{x})} D \psi (\vec{x}) \right] \Psi [\psi ,t][/tex]

where
[tex]D = -i \vec{\alpha}. \vec{\nabla} + m \beta [/tex]
is the Dirac operator.
So, Dirac wavefunctional (which maps points in the spinor space into numbers) is a solution to Grassmann functional differential equation which represents the Schrodinger wave equation in spinor field theory.

regards

sam

 

[jostpuur]

I'll first ask about notation.

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

[tex]\partial_i\Psi[/tex] or [tex]\partial_{x_i}\Psi[/tex].

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

[tex]\partial_x\Psi[/tex] or [tex]\partial_{\phi_x}\Psi[/tex].

Is this the same thing that [tex]\frac{\delta}{\delta \phi(x)}\Psi[/tex] means?

 

[samalkhaiat]

I'll first ask about notation.

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

[tex]\partial_i\Psi[/tex] or [tex]\partial_{x_i}\Psi[/tex].

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

[tex]\partial_x\Psi[/tex] or [tex]\partial_{\phi_x}\Psi[/tex].

Is this the same thing that [tex]\frac{\delta}{\delta \phi(x)}\Psi[/tex] means?

More or less, Yes.
A fuction f on spacetime maps points in spacetime to real or complex numbers. Now, let f be a point in a "scalar" function space, A. Let F be a functional on A. F maps poins in A into real or complex numbers;

[tex] F = F[f] \in \mathbb{C} \ \mbox{or} \ \mathbb{R}[/tex]

The simplest functional is

[tex]F[f] = \int f(x) dx [/tex]

As we move from point to point in the functional space A, the value of a functional, F[f], will change. The rate of change of F with respect to a change in the function f is expressed as the functional derivative of F. It is formally defined by;

[tex]\frac{\delta}{\delta f} F[f] = \lim_{a \rightarrow 0} \frac{F[f + a \delta] - F[f]}{a}[/tex]

where [itex]\delta[/itex] in the argument of F is Dirac delta function.


regards

sam

 

[jostpuur]

[tex]i \frac{\partial}{\partial t} \Psi [ \psi ,t] = \int dx^{3} \left[ \frac{\delta}{\delta \psi (\vec{x})} D \psi (\vec{x}) \right] \Psi [\psi ,t][/tex]

where
[tex]D = -i \vec{\alpha}. \vec{\nabla} + m \beta [/tex]
is the Dirac operator.

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

 

[samalkhaiat]

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

Ok, in the spinor field, Dirac Hamiltonian is given by

[tex]H(\psi ,\pi) = \int dx^{3} \psi^{\dagger}D \psi[/tex]

where D is the Dirac operator and [tex]\pi[/tex] is the field conjugate to [tex]\psi[/tex] (momentum);

[tex]\pi = i \psi^{\dagger}[/tex]

In the "coordinate" representation

[tex]\pi \rightarrow i \frac{\delta}{\delta \psi}[/tex]

Putting these in Schroginger equation:

[tex]\partial_{t} \Psi[\psi ,t] = H(\pi , \psi) \Psi[\psi ,t][/tex]

gives us the Grassmann functional differential equation.


sam

 

[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

 

[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
See also Sec. 4 of the paper above for an alternative.