# What is the meaning of uncertainty principle in QTF?

1. Jul 6, 2011

### ndung200790

Please teach me this:
In Quantum Mechanics the Heizenbeg principle holds for one particle.In QTF Theory,the creation and annihilation operators corresonding to coefficients of Fourier components of transformation,so it seem that each quantum(particle) has a definite momentum.Then it seem that the uncertainty principle holds for inter-particles(many particles),but not for single particle.If it is true,where is the role of Quantum Mechanics in Quantum Field Theory?
Thank you very much in advance.

2. Jul 7, 2011

### ndung200790

It seem that because the Identity Principle,we ought to not pose the ''distinguishing'' between one particle(in QM) and many particles(in QTF Theory).Is that correct?

3. Jul 7, 2011

### Staff: Mentor

Did you mean QFT? Quantum Field Theory?

4. Jul 7, 2011

### Dickfore

QFT is a theory of fields as physical systems. A field has (an uncountably) infinite number of degrees of freedom labeled by the space-time points $x = (ct, \mathbf{r})$. Then, if you have a Lagrangian density $\mathcal{L}$, you can find the canonical momentum density:
$$\pi(\mathbf{r}, t) = \frac{\partial \mathcal{L}}{\partial (\partial_{t}\varphi(\mathbf{r}, t))}$$
The quantum part of the theory are the canonical commutation relations:
$$\begin{array}{lcl} [\varphi(\mathbf{r}, t), \varphi(\mathbf{r}', t)]_{\mp} & = & 0 \\ [\varphi(\mathbf{r}, t), \pi(\mathbf{r}', t)]_{\mp} & = & i \, \hbar \, \delta^{(3)}(\mathbf{r} - \mathbf{r}') \\ [\pi(\mathbf{r}, t), \pi(\mathbf{r}', t)]_{\mp} & = & 0 \end{array}$$
taking the commutator or anticommutator depending on whether the fields correspond to integer or half-integer spin (see Spin-statistics Theorem). Notice that the field operators entering in the above relations are taken at the same instant in time. In this regard, the formulation is apparently not relativistically invariant, but the requirement for locality and the special properties of the fields under Lorentz transformations actually make it so.

5. Jul 7, 2011

### ndung200790

So I guess you mean the space coordinate and momentum in Quantum Field Theory simultaneously having the definite values?Is it correct you mean that?

6. Jul 7, 2011

### Dickfore

no, in QFT the role of generalized coordinates is taken by the field ($\varphi(x)$) and the space-time 4-vector x plays the role of a (continuous) label. Momentum is defined just like in mechanics of systems with finite number of degrees of freedom.

7. Jul 7, 2011

### ndung200790

I think that even inner the QTF Theory,there are some ''average'' value,e.g the momentum p,the energy epsilon of the particle(quantum).

8. Jul 7, 2011

### ndung200790

Then,are there any uncertainty relation between momentum(value) and field(value(not operator))?

9. Jul 7, 2011

### Dickfore

particles in qft are identified as specific excitations of the field that carry a certain amount of momentum and energy. These excitations are actually the different Fourier modes that are solutions to the equations of (a non-interacting) qft.

The basic problem in qft is the following: Given a particular excitation prepared in the distant past (corresponding to incoming particles), what is the transition amplitude for a particular arrangement of (not necessarily the same) particles coming out in the distant future. This is an idealized scattering experiment.

The square of the modulus of the transition amplitude, plus some extra factors depending on the phase-space volume give the cross section for the particular process. In this way, QFT is probabilistic because there are many outcomes to a scattering experiment.

10. Jul 7, 2011

### ndung200790

Thanks very much for your kind and useful helping.

11. Jul 7, 2011

### dx

In quantum field theory, unambiguous meaning can be attached only to spacetime integrals of the idealized field components. Thus when comparing two measurements of field averages over different spacetime regions, we cannot speak unambiguously about the temporal sequence of the measurements and therefore in the discussion of measurement possibilieties in field theory essentially new considerations are involved. In the investigations of Landau and Pierls on the measurability of electromagnetic field quantities, they employed point charges as test bodies and therefore were lead to incorrect conclusions. In fact, only measuring bodies with a charge distribution of finite extent should be considered in the testing of the quantum electrodynamic formalism, since every well-defined statement of this formalism refers to averages of field components over finite spacetime regions.

12. Jul 7, 2011

### ndung200790

QFT mean Quantum Field Theory

13. Jul 7, 2011

### ndung200790

Then,I think that there is a difference between quantum and particle.The quantum has definite 4-momentum,but the particle may or may not have definite 4-momentum.The only particle may have definite space-time coordinate,but quantum corresponding to a Fourier component of field.May be the quantum notion is ''narrower'' than particle notion.Quantum Mechanics use ''particle-wave'' notion but Quantum Field Theory use ''quantum-wave(or field)'' notion.Is that correct?

14. Jul 7, 2011

### ndung200790

It seem that one particle in QTF Theory might be ''involve'' some or many quantum.E.g a definite state of electron might involve some ''quantum''.

15. Jul 7, 2011

### Polyrhythmic

In QFT, particles are represented by fields, which can have quantized excitations. I'm not sure what you mean by a distinction between "quantum and particle".

16. Jul 7, 2011

### ndung200790

If quantum and particle are the same,why they have the definite 4-momentum which is momentum of Fourier component,because in general a particle in a superposition state has not a certain value of 4-momentum?

17. Jul 8, 2011

### ndung200790

I think that the reation and annihilation process of particle(quantum) has probability characteristic.The process changes the Fock states from one to another.If a quantum(particle) has a great momentum,then the fluctuation of momentum in the Fock state ''space'' is large,so following Quantum Mechanics(many body) the coordinate r of particle is definite(Heizenberg principle).Is that correct?
By the way,it seem that the Fourier component of field can not be considered as the wave function of particle(corresponding Fourier component quantum )

18. Jul 8, 2011

### ndung200790

By the way,please teach me what is the difference between the value(not operator) of quantum field and wave function?

19. Jul 9, 2011

### A. Neumaier

The Fourier components of a quantum field are still operators.

Wave functions in QFT are either (in the functional representation of Phi^4 theory, say) functions of the field Phi, or in the more often used Fock representation sums over N of integrals over N-particle momentum states.

Measurable values of a quantum field are expectations of the field (at particular points in space-time). Also measurable are simple enough correlation functions, which are Fourier transforms of expectations of products of fields.

20. Jan 28, 2013

### DrViking

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