# How are the positions of particles described in QFT?

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.

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?
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.

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)

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.
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.

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)
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.

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Demystifier
Gold Member
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.
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.

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.

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

Hans de Vries
Gold Member
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.
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.

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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...
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.

Demystifier
Gold Member
Asking for the position of a particle below its Compton wavelength is a unphysical question.
And the Compton wavelength of the photon is ... ?

Demystifier
Gold Member
Those are not my words, but those of Sidney Coleman, made in his 'legendary' lectures on QFT.
And other people uncritically repeat his words.

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

Hans de Vries
Gold Member
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.

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

Demystifier
Gold Member
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.
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?

Hans de Vries
Gold Member
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
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

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Hans de Vries
Gold Member
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?
This is the massless-limit and massless particles don't have a time-evolution because they
move at the speed of light.

Regards, Hans

Hi Hans,

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

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

Hans de Vries
Gold Member
Hi Hans,

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

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

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.
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

the $\psi$ is not an eigenfunction

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

which is valid only for eigenfunctions of the Hamiltonian.

Eugene.

Hans de Vries
Gold Member

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

which is valid only for eigenfunctions of the Hamiltonian.

Eugene.
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

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.
I respectfully disagree. If equation

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

is valid at t=0, then

$$H \psi \ =\ \pm \ mc^2 \sqrt{ 1 - \frac{r_o^2}{r_x^2}}\ \ \ \psi$$

which means that $\psi$ is an eigenstate of the Hamiltonian and its time evolution is given by the simple phase factor $\exp (-\frac{i}{\hbar} at)$, where

$$a =\ \pm \ mc^2 \sqrt{ 1 - \frac{r_o^2}{r_x^2}}$$

is the eigenvalue of $H$.

Eugene.

Hans de Vries
Gold Member
which means that $\psi$ is an eigenstate of the Hamiltonian and its time evolution is given by the simple phase factor $\exp (-\frac{i}{\hbar} at)$, where

$$a =\ \pm \ mc^2 \sqrt{ 1 - \frac{r_o^2}{r_x^2}}$$

is the eigenvalue of $H$.
Eugene.
You are almost correct. The time dependent wave function:

$$\psi(r,t)\ =\ \frac{1}{r}\ \exp{\left(-\frac{r}{r_x}\ \pm\ i \frac{mc^2}{\hbar}\ t\ \sqrt{ 1 - \frac{r_o^2}{r_x^2}}\right)}$$

would be a localized non-spreading stationary solution of the equation
if it wasn't for the single point in the middle (the singularity) where it
doesn't hold... When rx gets below the Compton radius one could call
this wave function the "Big Bang" function due to its exponential
growth nature. In practice, any more physical cut-off instead of a
mathematical singularity would stop the catastrophic process...

Regards, Hans

Hi Hans,

now I see where is the source of the problem

Using the wave function :

$$\psi\ =\ \frac{e^{-r/r_x}}{r}$$...........(1)

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:

$$\nabla^2\psi\ =\ \frac{1}{r_x^2}\ \frac{e^{-r/r_x}}{r}$$...........(2)
The Laplacian is indeed the one you wrote in (2) everywhere, except in the origin, where you should exercise extra care due to the divergent factor $1/r$. If your equation (2) was correct, then (1) would be a stationary solution for the free non-relativistic particle. In fact, it is not. It is actually a stationary bound state for the attractive delta-function potential centered in the origin. You can convince yourself that this is, indeed, the case by taking bound state solutions for a spherical well potential and study their limits when the radius of the well goes to zero and its depth goes to infinity.

Eugene.

Hans de Vries
Gold Member
Hi Hans,

now I see where is the source of the problem
The Laplacian is indeed the one you wrote in (2) everywhere, except in the origin, where you should exercise extra care due to the divergent factor $1/r$. If your equation (2) was correct, then (1) would be a stationary solution for the free non-relativistic particle.
Yes, that's indeed my point.

In fact, it is not. It is actually a stationary bound state for the attractive delta-function potential centered in the origin. You can convince yourself that this is, indeed, the case by taking bound state solutions for a spherical well potential and study their limits when the radius of the well goes to zero and its depth goes to infinity.
Did you check this? Infinitesimal deep potential wells have generally zero
probability for a particle to leak outside.

Strong localization leads to states with negative energy frequencies
in the case of Schrödinder's equation when the contribution of
$$\nabla^2\psi$$ due to high (positive) second order derivatives becomes larger as
the rest mass energy. You can simply read this directly from the equation
itself.

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

Regards, Hans

Strong localization leads to states with negative energy frequencies
in the case of Schrödinder's equation when the contribution of
$$\nabla^2\psi$$ due to high (positive) second order derivatives becomes larger as
the rest mass energy. You can simply read this directly from the equation
itself.

$$-i \hbar \frac{\partial \Psi (\mathbf{r},t)}{\partial t}\ =\ \frac{\hbar^2 c^2}{2m} \nabla^2\ \Psi (\mathbf{r},t)\ -\ U(\mathbf{r})\Psi (\mathbf{r},t)$$
Localization of relativistic particles in strong attractive potentials may lead to some peculiar (though not necessarily unphysical) effects. For example, imagine an atomic nucleus with charge Z and corresponding attractive Coulomb potential for electrons. The lowest state available for electrons has energy E. If you increase the nuclear charge, the energy E would increase, and at some point (I think this happens at Z > 137) this energy becomes larger than the energy $2mc^2$ required to create one electron-positron pair. As far as I know, such conditions have not been achieved experimentally. However, the current understanding about what happens next is this: The electron-positron pair, indeed, gets created. The electron gets attached to the nucleus, thus effectively reducing its charge to Z-1, and the positron is repelled to infinity.

Note that such pair creation requires a strong external potential. However, if are you simply localizing a free particle in empty space, then there is no reason to expect creation of pairs.

Eugene.