# How are the positions of particles described in QFT?

1. Oct 2, 2007

### vcdfrexzaswq

If I understand correctly positions of particles cannot have exact values in QFT, there are no eigenvectors of position (right?). But the positions of particles must correspond approximately to some state in QFT because position is meaningful in QM and classical physics, and QFT is supposed to explain these.

Has an example, if a fremion is located approximately at some position, can you describe it has superposition’s of the Dirac field?

Sorry if I worded things wrong.

2. Oct 2, 2007

### meopemuk

Similar questions were discussed in several recent threads, for example in

Here is my take on it: Traditional QFT does not provide any meaningful way to discuss particle positions. QFT is normally formulated in the "bare" particle representation, and these "bare" particles have very distant relationship to physical particles observed in experiments. In particular, Dirac (and other quantum) fields have no connection to particle wave functions, i.e., probability amplitudes.

There is a way to reformulate QFT in the so-called "dressed (or physical) particle" representation, where Hermitean operators of particle positions and corresponding wave functions have precise definitions and clear physical meaning. This representation provides a link from QFT to ordinary non-relativistic quantum mechanics and classical physics.

Eugene.

3. Oct 3, 2007

### Ratzinger

When going below Compton wavelength of a particle, energy uncertainty is big enough that particle creation occurs. So asking for the position of a particle in relativistic quantum theory does not make sense.

In QFT position is no longer an observable, but becomes a parameter to operator-valued fields that act on many-particle states. (Momentum operators and it's eigenstates)

4. Oct 3, 2007

### meopemuk

This argument is often made in books and articles, however, it does not look convincing to me. I can agree that localizing a particle below its Compton wavelength implies large energy uncertainty. This follows from the fact that operators of position and energy do not commute. But I don't agree that sharp localization implies particle creation. The operator of position (Newton-Wigner) commutes with the particle number operator. So, there should be no problem in defining a well-localized state of a single physical particle in QFT.

If this point of view is accepted then there is no way to obtain ordinary non-relativistic quantum mechanics and classical mechanics (in which particle position observables are well-defined) as limiting cases of QFT. I think that arguments $\mathbf{r}$ of QFT quantum fields $\psi (\mathbf{r},t)$ have no relationship to experimentally measured positions. Instead, particle positions should be identified with (eigenvalues of) Newton-Wigner operators defined in each n-particle sector of the Fock space.

Eugene.

Last edited: Oct 4, 2007
5. Oct 4, 2007

### Demystifier

First, particle creation occurs due to interactions, not due to energy uncertainty.
Second, the Compton wavelength of a photon is infinite, yet we measure positions of photons with reasonably well accuracies.

6. Oct 4, 2007

### Ratzinger

More precise:localizing ( by scattering, putting in a box) a particle below Compton wavelength leads to particle creation.

Asking for the position of a particle below its Compton wavelength is a unphysical question.

Those are not my words, but those of Sidney Coleman, made in his 'legendary' lectures on QFT.

7. Oct 4, 2007

### meopemuk

Particle position is arguably the most evident and easily measurable observable in macroscopic physics. If you deny the existence of the position observable, how are you going to connect QFT to (its limiting cases) QM and classical mechanics, where position plays a prominent role? I understand that one doesn't need position observable in QFT S-matrix calculations. However, if you decided to go beyond the S-matrix and study the time evolution of physical states, I don't think you can manage without position observable and wave functions in the position representation.

Eugene.

8. Oct 4, 2007

### Hans de Vries

A wave function smaller as the Compton radius ceases to be a wave function in the
sense we know it. It stops oscillating and grows exponentially until it's larger as the
Compton radius again. At least that's what you get when you insert this:

$$\psi\ =\ \frac{e^{-r/r_x}}{r}$$

In the Klein Gordon equation, where $r_o=\hbar c/m$ is the Compton radius.

$$\frac{\partial^2 \psi}{\partial t^2}\ =\ \left(-\frac{1}{r_o^2} + \frac{1}{r_x^2} \right)\psi$$

According to Schrödinger's equation it becomes a negative energy (anti-) particle...

Regards, Hans.

Last edited: Oct 4, 2007
9. Oct 4, 2007

### meopemuk

I don't see anything wrong in this wave function and its time evolution described by the Schroedinger's equation

$$-i \hbar \frac{\partial \psi (\mathbf{r},t)}{\partial t}\ =\ \sqrt{m^2c^4 - \hbar^2 c^2 \nabla^2}\psi (\mathbf{r},t)$$

which is a position-space representation of the time evolution equation for states

$$|\psi(t) \rangle = \exp(\frac{i}{\hbar} \sqrt{m^2c^4 + c^2 \hat {\mathbf{p}}^2})|\psi (0) \rangle$$

Could you please explain why you think this time evolution leads to antiparticles?

Eugene.

10. Oct 5, 2007

### Demystifier

And the Compton wavelength of the photon is ... ?

11. Oct 5, 2007

### Demystifier

And other people uncritically repeat his words.

12. Oct 5, 2007

### Demystifier

The fact that a question is not answered by a particular theory (in this case, the standard formulation of QFT) does not make this question unphysical. Unless you postulate that this theory cannot be improved, even in principle.

13. Oct 5, 2007

### Hans de Vries

This is not the Schrödinger equation but a square-rooted version of the Klein
Gordon equation and it leads to the same conclusion:

A wave function smaller as the Compton radius ceases to be a wave function in the
sense we know it. It stops oscillating and grows exponentially until it's larger as the

So, the exponential growing of the wave function can be related to particle creation
in the way Sidney Coleman did.

Regards, Hans

14. Oct 5, 2007

### Demystifier

So I will repeat, as long as other people here try to ignore this. What is the Compton wavelength of the photon? Infinite, of course. So, does it mean that it does not make sense to have 1 photon in a large box?

15. Oct 5, 2007

### Hans de Vries

To back this up mathematically: Your square rooted Klein Gorden equation:

$$-i \hbar \frac{\partial \psi (\mathbf{r},t)}{\partial t}\ =\ \sqrt{m^2c^4 - \hbar^2 c^2 \nabla^2}\psi (\mathbf{r},t)$$

Can be written as:

$$i \hbar \frac{\partial}{\partial t} \ \psi \ =\ \pm \ mc^2\left\{ 1 - \sum_{n=1}^\infty \frac{ (2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{\hbar^2}{m^2 c^2}\ \right)^{n} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right)^{n}\right\} \psi$$

Using the wave function :

$$\psi\ =\ \frac{e^{-r/r_x}}{r}$$

Which becomes "smaller as the Compton radius"" when r_x becomes smaller as r_o.
Where r_o is the Compton radius. The Laplacian of this wave function is:

$$\nabla^2\psi\ =\ \frac{1}{r_x^2}\ \frac{e^{-r/r_x}}{r}$$

Substituting this in the series expansion gives:

$$i \hbar \frac{\partial}{\partial t} \ \psi \ =\ \pm \ mc^2\left\{ 1 - \sum_{n=1}^\infty \frac{ (2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{r_o^2}{r_x^2}\ \right)^{n}\right\} \psi$$

Which is the series expansion of:

$$i \hbar \frac{\partial}{\partial t} \ \psi \ =\ \pm \ mc^2 \sqrt{ 1 - \frac{r_o^2}{r_x^2}}\ \ \ \psi$$

And this is just the square root of my result in #8 as expected. The square root
becomes imaginary and time evolution becomes exponential grow. (The series
diverges below the compton radius )

Regards, Hans

Last edited: Oct 5, 2007
16. Oct 5, 2007

### Hans de Vries

This is the massless-limit and massless particles don't have a time-evolution because they
move at the speed of light.

Regards, Hans

17. Oct 5, 2007

### meopemuk

Hi Hans,

I am not sure where exactly, but you probably made a mistake in calculations. Your equation

cannot be valid, because it implies

$$H \psi (\mathbf{r}) =\ \pm \ mc^2 \sqrt{ 1 - \frac{r_o^2}{r_x^2}}\ \ \psi (\mathbf{r})$$.............(1)

where

$$H = \sqrt{m^2c^4 + p^2c^2}$$...........(2)

is the free particle Hamiltonian. Eq. (1) implies that the function $\psi (\mathbf{r})$ is an eigenfunction of this Hamiltonian with eigenvalue that can be negative, or even imaginary. However, operator (2) is Hermitean and positively-definite. So, all its eigenvalues must be positive. This is a contradiction.

Eugene.

18. Oct 5, 2007

### Hans de Vries

The math is correct but the $\psi$ is not an eigenfunction because it's not
stationary, on the contrary, instead of oscillating it will grow exponently .
This is what you get if you confine the wave function to a space smaller as the

Regards, Hans

19. Oct 5, 2007

### meopemuk

$$i \hbar \frac{\partial}{\partial t} \ \psi \ =\ \pm \ mc^2 \sqrt{ 1 - \frac{r_o^2}{r_x^2}}\ \ \ \psi$$

which is valid only for eigenfunctions of the Hamiltonian.

Eugene.

20. Oct 5, 2007

### Hans de Vries

The "=" in the equation here applies only at "t=0". (psi is an eigenfunction when the
equation is valid at all t). The equation is a snapshot: Compress the wave function
to within the Compton radius and then look at its basic time-evolution behavior just
after you let it go.

Regards, Hans