How observation leads to wavefunction collapse?

[marlon]

a single electron does not create an interference pattern

Sorry to bumb in this late but i think your statement is a bit inaccurate. It is indeed true we need to redo several times the double slit experiment to actually observe the interference pattern. But, 1 electron is interfering with itself. More specifically, the interference happens between all the possible paths when going from source to detector. So, not observing the interference pattern after 1 single measurement does NOT imply that the electron is NOT interfering !

Just my 2 $'s


marlon

 

[meopemuk]

If the probability is not 100%,I don't see why it should follow a different law/transformation.

I agree with Fra on this point:

... I agree on the fact that probabilities can't just be transformed like a lorentz scalar. ... probability is not an objective thing, it's not an object that everyone agress upong. It's intrinsically relative to the observer. What is probably for me, need not be probably to you for the simple reason that we devise our ensembles differently.

I think that your transformation law $\psi(x) \to \psi(\Lambda x)$ is not obvious at all. If you think it is true, you should be able to prove it, rather that give a vague reference to analogies.

I asked you to complete your calculation in post #252 with the time coordinate included.

This is done most easily in the momentum representation. If $\psi(\mathbf{p})$ is the wave function at time t=0 from the point of view of observer at rest O, then

$$e^{\frac{it}{\hbar} E_p}\psi(\mathbf{p})$$

is the wavefunction at time t from the point of view of O


$$\sqrt{\frac{E_{\Lambda p}}{E_p}}\psi(\Lambda \mathbf{p})$$

is the wavefunction from the point of view of moving observer O' at time t=0 (by her clock)

$$e^{\frac{it'}{\hbar} E_{\Lambda p}} \sqrt{\frac{E_{\Lambda p}}{E_p}}\psi(\Lambda \mathbf{p})$$

is the wavefunction seen by O' at time t' (by her clock). You can get position-space representations of these wavefunctions by making usual Fourier transformations. I don't think these integrals can be written in a nice analytical form.

The above formulas follow from the general fact that boost transformations are realized by unitary operators $\exp(-\frac{ic}{\hbar} K \theta)$, where $K$ is the generator of boosts; and time translations are realized by unitary operators $\exp(\frac{i}{\hbar} Ht)$, where $H$ is the Hamiltonian.

Anyway---if the wavefunction in one frame is $$\exp{\iota p_\mu x^\mu}$$(where $$\hbar=1$$),then you can see that the transformed wavefunction is going to be not much different.

This is true, but the wavefunction you wrote is just a specific case of a plane wave, i.e., a state with definite momentum, where our formulas coincide. I was talking about transformations of general wavefunctions.

It is possible to show (I will not do it here, but I am ready to reproduce calculations, if you are interested) that if we

1) take the classical limit of the above theory and consider trajectories of two non-interacting particles that intersect at the space-time point (x,t) from the point of view of observer O and
2) transform these trajectories according to commutators between the Newton-Wigner operator and generators of boost transformations $U_g$,

then

3) we will find that from the point of view of O', the intersection of trajectories occurs at point (x',t') which is related to (x,t) by usual Lorentz formulas (1) and (2).

Go ahead!

OK. To make things simple, I'll consider a one-dimensional problem in which the movement of the particles and the reference frame occur only along the x-axis. I will denote P, H, and K the generators of the Poincare group corresponding to space translations, time translations, and boosts, respectively. In the case of spinless particle, the Newton-Wigner position operator is $R = -c^2 \frac{K}{H}$ (In fact, the correct quantum expression is $R = -\frac{c^2}{2} (KH^{-1} + H^{-1}K)$, however, in the end we will take the classical limit anyway, so it is justified to ignore the non-commutativity of operators here.)

From the point of view of observer at rest O, the time dependence of R is

$$R(0,t) = \exp(\frac{i}{\hbar}Ht) R \exp(-\frac{i}{\hbar}Ht) = R + Vt$$

where $V = Pc^2/H$ is velocity (Calculations performed in this post can be easily done with the knowledge of commutators between P, H, K, R, and V. I'll skip the detailed calculations, but, if you like, I can justify all steps)

From the point of view of moving observer O', the position at t'=0 (where time t' is measured by the clock of observer O') is given by

$$R(\theta,0) = \exp(-\frac{ic}{\hbar}K \theta) R \exp(\frac{ic}{\hbar}K \theta) = \beta \frac{R}{\cosh \theta}$$

(Here I will use a convenient notation $\cosh \theta = (1-v^2/c^2)^{-1/2}$ and $\sinh \theta = v/c (1-v^2/c^2)^{-1/2}$ and
$\beta = (1 - Vv/c^2)^{-1}$, where v is the velocity of the reference frame O')

The time dependence of position from the point of view of O' is

$$R(\theta,t') = \exp(\frac{i}{\hbar}H't') R (\theta,0)\exp(-\frac{i}{\hbar}H't') = \beta (\frac{R}{\cosh \theta} + (V-v)t'} )$$ (1)

where

$$H' = \exp(-\frac{ic}{\hbar}K \theta) H \exp(\frac{ic}{\hbar}K \theta)$$

is the Hamiltonian in the reference frame O'.


Eq. (1) is the transformation of position obtained by quantum-mechanical Poincare-group analysis. On the other hand, standard Lorentz transformations of special relativity would predict

$$R(\theta,t') = R (0,t)\cosh \theta - ct \sinh \theta$$ (2)

$$t' = t \cosh \theta - \frac{R (0,t)}{c} \sinh \theta$$ (3)

Formulas (1) and (2) do not look identical. However, if we take their difference, replace t' by eq. (3) and use $R (0,t) = R + Vt$, we will get exactly zero.

This means that space-time coordinates of free classical particles transform exactly by Lorentz formulas within relativistic quantum mechanics. This also means that if there is a localized event defined as intersection of trajectories of two non-interacting particles, then space-time coordinates of this event will transform by Lorentz formulas as well.

One important remark. Here we obtained Lorentz transformation formulas, which are usually interpreted as an evidence of equivalence of space and time coordinates. However, there is no such symmetry in our approach: Position is described by a Hermitian operator, but time is just a classical parameter that labels reference frames.

Another important point is that by this method we will not be able to prove Lorentz formulas (2) and (3) for the point of collision if the two particles interact with each other.

Eugene.

 

[gptejms]

This is done most easily in the momentum representation. If $\psi(\mathbf{p})$ is the wave function at time t=0 from the point of view of observer at rest O, then

$$e^{\frac{it}{\hbar} E_p}\psi(\mathbf{p})$$... (1)

is the wavefunction at time t from the point of view of O


$$\sqrt{\frac{E_{\Lambda p}}{E_p}}\psi(\Lambda \mathbf{p})$$

is the wavefunction from the point of view of moving observer O' at time t=0 (by her clock)

$$e^{\frac{it'}{\hbar} E_{\Lambda p}} \sqrt{\frac{E_{\Lambda p}}{E_p}}\psi(\Lambda \mathbf{p})$$...(2)

is the wavefunction seen by O' at time t' (by her clock). You can get position-space representations of these wavefunctions by making usual Fourier transformations. I don't think these integrals can be written in a nice analytical form.

You have now transformed t to t',E to E'(in addition to p to p' which you were doing earlier)--now you add to this x to x' and you'll end up saying exactly what I am saying!

This is true, but the wavefunction you wrote is just a specific case of a plane wave, i.e., a state with definite momentum, where our formulas coincide. I was talking about transformations of general wavefunctions.

A general wavefunction is nothing but a sum of such plane waves, which is what you (would be) ultimately doing above,or in your post #252.

Formulas (1) and (2) do not look identical. However, if we take their difference, replace t' by eq. (3) and use $R (0,t) = R + Vt$, we will get exactly zero.

Another important point is that by this method we will not be able to prove Lorentz formulas (2) and (3) for the point of collision if the two particles interact with each other.

Eugene.

Interesting operator,but with limitations.

 

[meopemuk]

You have now transformed t to t',E to E'(in addition to p to p' which you were doing earlier)--now you add to this x to x' and you'll end up saying exactly what I am saying!


A general wavefunction is nothing but a sum of such plane waves, which is what you (would be) ultimately doing above,or in your post #252.

In order to find the position-space wavefunction for the moving observer O' at time t' one should first find eigenvectors $|\mathbf{r}, t' \rangle$ of the corresponding position operator

$$R(\theta,t') = \exp(\frac{i}{\hbar}H't') R (\theta,0)\exp(-\frac{i}{\hbar}H't') = \beta (\frac{R}{\cosh \theta} + (V-v)t'} )$$

$$R(\theta,t') |\mathbf{r}, t' \rangle = \mathbf{r}|\mathbf{r}, t' \rangle$$

and then take the inner product of the state vector $|\Psi \rangle$ with these eigenvectors

$$\psi (\mathbf{r}, t') = \langle \mathbf{r}, t' |\Psi \rangle$$ ... (1)



I haven't done this calculation, but I am pretty sure that the result will be different from your proposal $\psi (\Lambda x)$ , where $x \equiv (\mathbf{r}, t)$. One reason is this: if the state $|\Psi \rangle$ is localized from the point of view of observer O (this means that $\psi ( \mathbf{r}, 0)$ is zero everywhere except point $\mathbf{r}= 0$), then transformed wavefunction (1) is not localized (neither at t'=0 nor at any other time), but function $\psi (\Lambda x)$ is localized at t'=0.


Eugene.

 

[Hans de Vries]

In order to find the position-space wavefunction for the moving observer O' at time t' one should first find eigenvectors $|\mathbf{r}, t' \langle$ of the corresponding position operator.

Eugene,

When will you finally read those first two pages of Jun John Sakurai's chapter 3?It shows that the probability density transforms as the 0th component of the
four vector probability current density:

$$\frac{i}{2m}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

In this way you can account for the fact that the probability density goes
from one Lorentz contracted state to another:

$$\frac{E}{m}\psi(x) \to \frac{E'}{m}\psi(\Lambda x)$$

Where E, E' and m can be local densities in the most general case.Regards, Hans

 

[Hans de Vries]

Sakurai also proofs very simply the unitarity of the probability density of
the Klein Gordon equation.

Given the probability current density:$$s_\mu\ =\ \frac{i}{2m}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

We have the requirement for unitarity that the four divergence of this
vanishes. (The continuity relation)

$$\frac{\partial s_\mu}{\partial x_\mu}\ =\ \frac{i}{2m}\left( \frac{\partial \phi^*}{\partial x_\mu}\frac{\partial \phi}{\partial x_\mu} - (\Box\phi^*)\phi - \phi^*\Box\phi -\frac{\partial \phi^*}{\partial x_\mu}\frac{\partial \phi}{\partial x_\mu} \right)\ =\ 0$$

If phi obeys the Klein Gordon equation then we can replace this by:

$$\frac{\partial s_\mu}{\partial x_\mu}\ =\ \frac{i}{2m}\left( \frac{\partial \phi^*}{\partial x_\mu}\frac{\partial \phi}{\partial x_\mu} - m^2\phi^*\phi - \phi^*m^2\phi -\frac{\partial \phi^*}{\partial x_\mu}\frac{\partial \phi}{\partial x_\mu} \right)\ =\ 0$$

Which trivially vanishes.Regards, Hans

 

[jostpuur]

Eugene,

When will you finally read those first two pages of Jun John Sakurai's chapter 3?It shows that the probability density transforms as the 0th component of the
four vector probability current density:

$$\frac{i}{2m}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

In this way you can account for the fact that the probability density goes
from one Lorentz contracted state to another:

$$\frac{E}{m}\psi(x) \to \frac{E'}{m}\psi(\Lambda x)$$

Where E, E' and m can be local densities in the most general case.Regards, Hans

But this isn't probability density, is it? I haven't checked it with with some solution of the KG equation myself, but if I assume that Demystifier knows what he is talking about in his "QM Myths"-paper, there exists positive frequency solutions, for which this KG current density is negative (somewhere).

Demystifier, judging by your paper, and by responses to it that I saw here, I might guess that this

I was quite suprised when I noticed last night, that the Klein-Gordon equation actually does conserve probability, if only positive frequency solutions are considered. Are you aware of this?

(which I posted in the "Mathematical signature of electronmagnetic field and electron's deBroglie functions" thread) is not very well known. The calculation isn't even very difficult. For a given initial wavefunction $\phi(0,\boldsymbol{x})$, the positive frequency time evolution given by KG equation is

$$\phi(t,\boldsymbol{x}) = \int\frac{d^3p\;d^3x'}{(2\pi)^3} \phi(0,\boldsymbol{x}') e^{-i(E_{\boldsymbol{p}}t - \boldsymbol{p}\cdot (\boldsymbol{x}-\boldsymbol{x}'))}$$

(Actually I'm not sure about the sign convention, but if that is not the standard positive frequency solution, it probably isn't ruining anything important here.) Assume

$$\int d^3x\; |\phi(0,\boldsymbol{x})|^2 = 1$$

and using the usual delta function representations you can verify that

$$\int d^3x\; |\phi(t,\boldsymbol{x})|^2 = 1$$

is true for any fixed $t>0$. With the same calculation you can also show, that probability is conserved for negative frequency solutions. However, the linear combinations of positive and negative frequency solutions do not conserve the probability, and this is why Klein-Gordon equation does not conserve probability for arbitrary solutions, and is also why it is impossible to prove the conservation of probability out of Klein-Gordon equation alone.

meopemuk, I think you were ignoring the paradox I mentioned earlier. The solutions of your relativistic Shrodinger's equation

$$i\partial_t \psi(t,\boldsymbol{x}) = \sqrt{-\nabla^2 + m^2} \psi(t,\boldsymbol{x})$$

are also solutions of the Klein-Gordon equation. Since fixing the sign of the frequency of the solution of Klein-Gordon equation determines its solution uniquely, defining time evolution of a wave function with relativistic Shrodinger's equation is equivalent to defining it with the Klein-Gordon equation and demanding only positive frequency solutions to be accepted. The paradox stemmed from assumptions that Shrodinger's equation was conserving the probability and Klein-Gordon equation was not, simultaneously! But this one I solved quite quickly.

 

[Demystifier]

1. But this isn't probability density, is it? I haven't checked it with with some solution of the KG equation myself, but if I assume that Demystifier knows what he is talking about in his "QM Myths"-paper, there exists positive frequency solutions, for which this KG current density is negative (somewhere).

2. Demystifier, judging by your paper, and by responses to it that I saw here, I might guess that this

(which a posted in the "Mathematical signature of electronmagnetic field and electron's deBroglie functions" thread) is not very well known. The calculation isn't even very difficult. For a given initial wavefunction $\phi(0,\boldsymbol{x})$, the positive frequency time evolution given by KG equation is

$$\phi(t,\boldsymbol{x}) = \int\frac{d^3p\;d^3x'}{(2\pi)^3} \phi(0,\boldsymbol{x}') e^{-i(E_{\boldsymbol{p}}t - \boldsymbol{p}\cdot (\boldsymbol{x}-\boldsymbol{x}'))}$$

(Actually I'm not sure about the sign convention, but if that is not the standard positive frequency solution, it probably isn't ruining anything important here.) Assume

$$\int d^3x\; |\phi(0,\boldsymbol{x})|^2 = 1$$

and using the usual delta function representations you can verify that

$$\int d^3x\; |\phi(t,\boldsymbol{x})|^2 = 1$$

is true for any fixed $t>0$.

1. Yes, that is exactly what the problem is.

2. I do not think that you obtain a delta function. This is because energy E_p also depends on the 3-momentum p, so integration over d^3p is not so trivial. Check it once again.

 

[Hans de Vries]

But this isn't probability density, is it? I haven't checked it with with some solution of the KG equation myself, but if I assume that Demystifier knows what he is talking about in his "QM Myths"-paper, there exists positive frequency solutions, for which this KG current density is negative (somewhere).

You always end up with this expression. (Just read Sakurai chapter 3.1 and 3.5)

$$j_\mu\ =\ \frac{i}{2m}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

for the probability current density. One does so with the Klein Gordon equation,
with the Schrödinger equation after the prescribed substitution $\phi=\Psi \exp(-imc^2t/\hbar)$
and with the Dirac equation after the the spin current has been separated out
via the Gordon decomposition.Yes, The 0th component which represents the probability density is
negative in the case of anti particles. Here we arrive at the 1934
Pauli-Weisskopf interpretation of j as the charge current density.

Both the probabilistic interpretation and the Pauli-Weisskopf interpretation
are valid and have been proved in countless experiments. The 0th component
of j is conserved either as the probability or the electric charge.

The non conservation issues can occur if the initial wavefunction at t=0 is
not combined with the correct $\partial_t \psi$ at t=0 which is rather logical because
one then programs a violation of the continuity relation right into the
boundary conditions.


Regards, Hans

 

[jostpuur]

2. I do not think that you obtain a delta function. This is because energy E_p also depends on the 3-momentum p, so integration over d^3p is not so trivial. Check it once again.

So we want to calculate

$$\int d^3x\; \phi^*(t,\boldsymbol{x})\phi(t,\boldsymbol{x})$$

Substituting the general solutions to this gives

$$=\int\frac{d^3x\;d^3p\;d^3y\;d^3p'\;d^3y'}{(2\pi)^6} \phi^*(0,\boldsymbol{y}) e^{i(E_{\boldsymbol{p}}t - \boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y}))} \phi(0,\boldsymbol{y}') e^{-i(E_{\boldsymbol{p}'}t - \boldsymbol{p}'\cdot(\boldsymbol{x}-\boldsymbol{y}'))}$$

This can be rearranged to be

$$=\int\frac{d^3x\;d^3p\;d^3y\;d^3p'\;d^3y'}{(2\pi)^6} \phi^*(0,\boldsymbol{y})\phi(0,\boldsymbol{y}') e^{i(E_{\boldsymbol{p}} - E_{\boldsymbol{p}'})t} e^{i(\boldsymbol{p}\cdot\boldsymbol{y} - \boldsymbol{p}'\cdot\boldsymbol{y'})} e^{i(\boldsymbol{p}'-\boldsymbol{p})\cdot\boldsymbol{x}}$$

Integration of variable x can be carried out, and it gives $\delta^{3}(\boldsymbol{p}'-\boldsymbol{p})$. Then you can integrate over the variable p', and this will remove energy terms.

 

[jostpuur]

Yes, The 0th component which represents the probability density is
negative in the case of anti particles. Here we arrive at the 1934
Pauli-Weisskopf interpretation of j as the charge current density.

The fact that it can be negative also for the positive frequency solutions is the true problem.

 

[Hans de Vries]

The fact that it can be negative also for the positive frequency solutions is the true problem.

This is simply not true. It's never negative for a positive frequency solution.
It can only be negative there where there is a negative energy density.
This does never occur for free particle solutions.

It also stays positive in the case of an interacting electron in a very deep
potential well where -eV is larger as the rest mass energy of the electron.
You have to take the interaction into account as is done in Sakurai chapter
3.5 in the paragraph handling the Gordon decomposition. (see 3.204)

For the Klein Gordon equation you get:

$$j_\mu\ =\ \frac{i\hbar}{2m}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)\ -\ \frac{e}{mc}A_\mu\phi^*\phi$$

where j_0 is the probability density.

Sakurai shows that unitarity also holds in the case of interacting fields.


Regards, Hans

 

[jostpuur]

This is simply not true. It's never negative for a positive frequency solution.

All right. I tried a superposition of two plane waves in one dimension, and the Klein-Gordon density was still positive everywhere. Perhaps Demystifier explains this.

 

[meopemuk]

Eugene,

When will you finally read those first two pages of Jun John Sakurai's chapter 3?

Hi Hans,

Following your suggestion I went to the library and found Sakurai's book. I wasn't impressed. Like many other authors Sakurai takes a heuristic approach to relativistic quantum mechanics, like "deriving" KG equation by analogy from $E^2 = m^2c^4 + p^2c^2$. This kind of guessing was forgivable in 1927, when people didn't know exactly what quantum mechanics is about. But we are in 2007 now. I think we can demand a more rigorous derivation of our theories from fundamental axioms of relativity and quantum mechanics.

It shows that the probability density transforms as the 0th component of the
four vector probability current density:

$$\frac{i}{2m}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

There are certain laws of quantum mechanics that we are not allowed to violate. For example, if you want to find a wavefunction corresponding to a state vector $|\Psi \rangle$ you need to perform the following steps:

1. Specify a full set of mutually commuting operators ($F_1, F_2, F_3, \ldots$) in the Hilbert space

2. Find common eigenvectors $|f_1, f_2, f_3, \ldots \rangle$ of this set

3. Then the wave function is

$$\psi(f_1, f_2, f_3, \ldots) = \langle f_1, f_2, f_3, \ldots | \Psi \rangle$$

4. and the probability (density) is

$$|\psi(f_1, f_2, f_3, \ldots)|^2$$

The same rules are valid for position-space wavefunctions. Your "probability current densities" are not built by these rules. So, you'll have to do some extra work to justify that they can be interpreted as probability densities. The fact that they satisfy the continuity equation is not a proof.

Eugene.

 

[meopemuk]

Sakurai also proofs very simply the unitarity of the probability density of
the Klein Gordon equation.

Given the probability current density:


$$s_\mu\ =\ \frac{i}{2m}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

We have the requirement for unitarity that the four divergence of this
vanishes. (The continuity relation)

No, this is not the definition of unitarity in quantum mechanics. According to Wigner theorem, the preservation of probabilities requires that state vectors are transformed by unitary operators. If you want to be consistent with quantum mechanics, you should prove that your "four divergence" definition agrees with Wigner theorem. Otherwise, you are not doing quantum mechanics. This is not a crime, of course. Anybody is allowed to look for alternatives to QM. But, at least, you should acknowledge that that's what you are doing.

Eugene.

 

[Hans de Vries]

Hi Hans,

Following your suggestion I went to the library and found Sakurai's book. I wasn't impressed.

Sakurai work is a classic and represents the mainstream ideas in relativistic
quantum mechanics.

There are certain laws of quantum mechanics that we are not allowed to violate. For example,
4. and the probability (density) is

$$|\psi(f_1, f_2, f_3, \ldots)|^2$$

The relativistic expression does just this! Just use the prescribed substitution
$\phi=\Psi \exp(-imc^2t/\hbar)$ in the relativistic expression: (explicitly expressing h and c)

$$j_\mu\ =\ \frac{i\hbar}{2mc^2}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

And you see that it becomes the classic expression for the Schrödinger
equation. The probability density HAS to be the 0th component of a four-
vector probability current density TO BE ABLE to hold under Lorentz
transformation.


Regards, Hans

 

[Hans de Vries]

No, this is not the definition of unitarity in quantum mechanics. According to Wigner theorem, the preservation of probabilities requires that state vectors are transformed by unitary operators. If you want to be consistent with quantum mechanics, you should prove that your "four divergence" definition agrees with Wigner theorem. Otherwise, you are not doing quantum mechanics. This is not a crime, of course. Anybody is allowed to look for alternatives to QM. But, at least, you should acknowledge that that's what you are doing.

Eugene.

This is a well established mainstream expression in relativistic quantum
mechanics!

I'm sure Wigner did agree with this and not with your ideas.Regards, Hans

 

[jostpuur]

The probability density HAS to be the 0th component of a four-
vector probability current density TO BE ABLE to hold under Lorentz
transformation.

Do you know for sure if the total Klein-Gordon charge remains constant in Lorentz transformations?

On the other hand, do you know about the same question concerning quantity
$$\int d^3x |\phi|^2$$
Does the normalization change in Lorentz transformations?

I cannot see easily why the probability density should be a first component of the four vector, instead of being scalar. Can it be proven with some calculation reasonably?

Hopefully Hans isn't already answering (I'm editing this here:). I can see that probability current should be a four vector so that the continuity equation gets satisfied manifestly in all frames. However, I just showed that also the normalization of $|\phi|^2$ remains constant in positive frequency time evolution, so I cannot see what's the problem with such scalar probability density.

In my previous questions I started thinking about the total charge and normalization in Lorentz transformations. If the discussion wasn't about then in the first place, then my comments were perhaps a bit confusing.

 

[meopemuk]

Just use the prescribed substitution
$\phi=\Psi \exp(-imc^2t/\hbar)$ in the relativistic expression: (explicitly expressing h and c)

$$j_\mu\ =\ \frac{i\hbar}{2mc^2}\left( \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right)$$

And you see that it becomes the classic expression for the Schrödinger
equation.

The time dependence $\phi=\Psi \exp(-imc^2t/\hbar)$ is characteristic for eigenfunctions of the energy operator, for which the probability density is time-independent. This time dependence cannot be valid in the general case.

The probability density HAS to be the 0th component of a four-
vector probability current density TO BE ABLE to hold under Lorentz
transformation.

Can you define the exact meaning of "to hold under Lorentz transformation"?
Can you prove your statement as a theorem? What would be your axioms in this case?

Eugene.

 

[meopemuk]

This is a well established mainstream expression in relativistic quantum
mechanics! :

Hi Hans,

I agree that what you wrote is "mainstream". I disagree that it is "well-established". Textbooks usually justify KG equation, "probability current densities", etc. by handwavings and vague analogies. Those who complain will be told that one-particle relativistic quantum mechanics does not make sense anyway, and should be replaced by the full-blown QFT approach. Or that $\mathbf{r}$ should not be interpreted as position, because position is not measurable in the relativistic world anyway. There are many ways to weasel out of hard questions. That's why I am insisting on an axiomatic approach.

Eugene.

 

[meopemuk]

Do you know for sure if the total Klein-Gordon charge remains constant in Lorentz transformations?

On the other hand, do you know about the same question concerning quantity
$$\int d^3x |\phi|^2$$
Does the normalization change in Lorentz transformations?

In Wigner's approach to wavefunctions and their transformations, the question of preservation of probabilities is trivial. Total probabilities are preserved with respect to time translations, boosts, and other Poincare transformations. This is true for wavefunctions written in the position or momentum or any other basis. This condition is explicitly satisfied, because wavefunction transformations are represented as unitary representations of the Poincare group.

Eugene.

 

[jostpuur]

In Wigner's approach to wavefunctions and their transformations, the question of preservation of probabilities is trivial. Total probabilities are preseved with respect to time translations, boosts, and other Poincare transformations. This is true for wavefunctions written in the position or momentum or any other basis. This condition is explicitly satisfied, because wavefunction transformations are represented as unitary representations of the Poincare group.

Eugene.

I believe this abstract way is valuable, but I still prefer doing things in a position representation too, just to make sure that the abstract principles are working. I'll probably return to this matter later, if I get something calculated.

At the moment I'm slightly confused about how there seems to be two similar conserved quantities in the Klein-Gordon field. $j^{\mu}$ and $|\phi|^2$.

I'm waiting eagerly to hear Demystifier's explanations about how $j^0$ can get negative, indicating incompatibility with a probability interpretation, and on the other hand Hans' explanations about why $|\phi|^2$, being scalar (and not a component of a four vector), would lead into problems with Lorentz transformations.

And I'm also interested to hear any opinions concerning the calculation that shows how $\int d^3x\;|\phi|^2$ is remaining constant. Since, I believe, it does surprise most its readers. Unless somebody soon points out some fatal flaw in it.

 

[Hans de Vries]

and on the other hand Hans' explanations about why $|\phi|^2$, being scalar (and not a component of a four vector), would lead into problems with Lorentz transformations.

The probability density has to transform like the 0th component of a
4-vector. It has to transform like Energy does:

$$\frac{E}{m}j_0(x) \to \frac{E'}{m}j_0(\Lambda x)$$

The wave function will be differently Lorentz contracted when going from
one reference frame to another. If you don't scale the values then the
integral over space changes by a factor $1/\gamma \rightarrow 1/\gamma'$

The factors E/m above correct the scaling for Lorentz contraction. For
example: If E/m=2 then you know that the wave function is Lorentz
contracted by a factor 2.

The Lorentz contraction depends on the reference frame AND the speed
of the wave function. This is where the 4-vector comes into play.Regards, Hans

 

[Hans de Vries]

In order to appreciate why the expression for the probability density
below transforms like Energy does,

$$j_0\ =\ \frac{i\hbar}{2m}\left( \phi^*\frac{\partial \phi}{\partial t} - \frac{\partial \phi^*}{\partial t}\phi \right)$$

and to see why it generates the required factor E/m, one can always
write the local instantaneous behavior of any wave function like this:

$$\phi\ \ \ =\ a\ e^{-iEt+bt}$$
$$\phi^*\ =\ a\ e^{+iEt+bt}$$

That is, it changes phase proportional to E and magnitude proportional to b,
while $a$ is a constant. Inserting this into the expression for the probability
density gives the required result while eliminating b:

$$j_0\ =\ \frac{E}{m}\ a$$

Again, if E/m is for instance 2, then we know that the wave function is
Lorentz contracted by a factor 2 and that we must scale the probability
density values by a factor 2 in order to have the integral over space equal
to 1.Regards, Hans

 

[jostpuur]

If we for example modeled probability with some kind of dust of small particles, and thought that probability is greater there where the density is greater, then I would understand that the probability should be a four current, but I don't understand why precisely should we model probability like this.

I understand that four momentum $p^{\mu}$ and the derivative operator $\partial^{\mu}$ both transform as four vectors, but your explanation still seems to be, that "it must be a four vector because it must be a four vector" (with an additional information that it is mainstream). But since I particularly would like to know why the scalar probability density $|\phi|^2$ is unacceptable, that explanation is not yet enough.

 

[jostpuur]

continuity equation without four current

I once studied a two component complex Klein-Gordon field, which satisfies the usual transformations of spin-1/2, a little bit. It had very interesting currents. I'm not getting into detail of it, but what happened was that it had a 16-component current

$$j^{\mu\nu}$$

which transformed as 2-rank tensor, so that for each fixed $\mu$ an equation

$$\partial_{\nu}j^{\mu\nu} = 0$$

was true. And for example

$$j^{0\nu}$$

was then a conserving current, even though it did not transform as a four vector.

Having encountered this example, I can believe that there are also more complicated examples of continuity equation being true, without the current being a four current.

In fact I see no reason to believe, that $|\phi|^2$ would not be a first component of some four component object, that satisfies the continuity equation.

 

[Hans de Vries]

Probability Current Density with Electro Magnetic interactionThe free field Probability current density is, with $j_o$ as the probability density:

$$j_\mu\ =\ \frac{i\hbar}{2m}\left[ \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right]$$For the 4-vector to transform correctly under Lorentz transform we must
substitute the derivatives with the Covariant derivatives:

$$j_\mu\ =\ \frac{i\hbar}{2m}\left[ \phi^*\left(\frac{\partial }{\partial x_\mu}-ieA_\mu\right)\phi - \phi\left(\frac{\partial }{\partial x_\mu}+ieA_\mu\right)\phi^* \right]$$

Or:

$$j_\mu\ =\ \frac{i\hbar}{2m}\left[ \phi^*\frac{\partial \phi}{\partial x_\mu} - \frac{\partial \phi^*}{\partial x_\mu}\phi \right]\ -\ \frac{eA_\mu}{m}\ \phi^*\phi$$Note that the probability now stays positive even in the case of an
electron in a very deep potential well where -eV is larger as the
electron's restmass.

Most textbooks fail to give this expression. Omitting the interaction term
would lead to a negative probability density in the case given above,
which might be one of the reasons for the myth that the probability
density of the Klein Gordon equation can be negative for particles.
(and positive for anti particles)

The same is true for the unitarity. Omitting the interaction term makes
the Klein Gordon equation non unitary when interacting Electromagnetically.Regards, Hans

 

[jostpuur]

Can you prove that the current density is nonnegative? It seems obvious only for plane waves.

 

[meopemuk]

The probability density has to transform like the 0th component of a
4-vector. It has to transform like Energy does:

$$\frac{E}{m}\psi(x) \to \frac{E'}{m}\psi(\Lambda x)$$

The wave function will be differently Lorentz contracted when going from
one reference frame to another. If you don't scale the values then the
integral over space changes by a factor $1/\gamma \rightarrow 1/\gamma'$

The factors E/m above correct the scaling for Lorentz contraction. For
example: If E/m=2 then you know that the wave function is Lorentz
contracted by a factor 2.

The Lorentz contraction depends on the reference frame AND the speed
of the wave function. This is where the 4-vector comes into play.

Do you have a proof of your transformation formula? It looks strange that the wavefunction becomes dependent on both position (x) and momentum (p) variables. (Momentum enters there through $E = \sqrt{m^2c^4 + p^2c^2}$, if I'm not mistaken) Normally wavefunction arguments should be eigenvalues of a commuting set of operators, but x and p do not commute with each other.

Eugene.

 

[Hans de Vries]

Do you have a proof of your transformation formula?

Let me change $$\psi(x)$$ to $$j_0(x)$$ to avoid confusion between the wave function
itself and the probability density. Given that E/m=1 in the restframe we get:

$$j_0(x) \to \frac{E'}{m}j_0(\Lambda x)$$

So $$j_0(x)$$ is the probability density (in the particles rest frame). The particle
is Lorentz contracted in other reference frames by a factor of gamma = E'/m.
So we must scale the probability values with the same factor E'/m to get
an integral over space equal to one. That's logical isn't it?

So with the wave function itself transforming like:

$$\psi(x) \to \psi(\Lambda x)$$

And the textbook formula for the probability density:

$$j_0\ =\ \frac{i\hbar}{2m}\left( \psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi \right)$$

We do indeed have a probability density which transforms like the 0th
component of a 4-vector (as the textbooks say). This is most easily
shown as I did a few posts back by writing the local instanteneous
behavior of the wave function like this:

$$\psi\ \ \ =\ a\ e^{-iEt+bt}$$
$$\psi^*\ =\ a\ e^{+iEt+bt}$$

That is, it changes phase proportional to E and magnitude proportional to b,
while $a$ is a constant. Inserting this into the expression for the probability
density gives the required result while eliminating b:

$$j_0\ =\ \frac{E}{m}\ a$$Regards, Hans

 

[meopemuk]

No, I don't accept it as a proof. Just a few objections:

The particle
is Lorentz contracted in other reference frames by a factor of gamma = E'/m.

I think that length contraction should be derived as a consequence of boost transformations of particle wavefunctions, not the other way around.

So with the wave function itself transforming like:

$$\psi(x) \to \psi(\Lambda x)$$

How do you prove this transformation law? This is, actually, the central point of our disagreements.

And the textbook formula for the probability density:

$$j_0\ =\ \frac{i\hbar}{2m}\left( \psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi \right)$$

There is another textbook formula for the probability density $|\psi|^2$, which makes more sense to me. My formula follows from fundamental laws of quantum mechanics. I don't know where your formula comes from.

Regards.
Eugene.

 

[Hans de Vries]

There is another textbook formula for the probability density $|\psi|^2$, which makes more sense to me. My formula follows from fundamental laws of quantum mechanics.

The formula you give is the non-relativistic version representing the limit
case of the fully relativistic formula.

I don't know where your formula comes from.

Regards.
Eugene.

As you very well know. These formula's can be found in all textbooks of
relativistic quantum mechanics. You should not proclaim that these
formula's do not exist since this is misleading for the students visiting
physicsforums.com and leads to confusion as has happened repeatedly
here.

If you do not agree with the formula's in the mainstream textbooks then
you should make it explicitly clear that this is your personal opinion/theory
and not proclaim your own personal opinion/theory as a given and common
accepted fact.

This is a (required) courtesy to the learning students and readers which
are visiting and using physicsforums.com Regards, Hans.

 

[meopemuk]

And the textbook formula for the probability density:

$$j_0\ =\ \frac{i\hbar}{2m}\left( \psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi \right)$$

I don't know where your formula comes from.

As you very well know. These formula's can be found in all textbooks of
relativistic quantum mechanics. You should not proclaim that these
formula's do not exist since this is misleading for the students visiting
physicsforums.com and leads to confusion as has happened repeatedly
here.

If you do not agree with the formula's in the mainstream textbooks then
you should make it explicitly clear that this is your personal opinion/theory
and not proclaim your own personal opinion/theory as a given and common
accepted fact.

This is a (required) courtesy to the learning students and readers which
are visiting and using physicsforums.com

Hi Hans,

please don't take it personally. I apologize for being so tense. Of course, I know that this formula is written in many textbooks. However, you would probably agree with me that textbooks don't do a good job in explaining the roots of this formula, and how it relates to the laws of quantum mechanics. I thought that we could try to go a bit deeper than simply cite textbooks.

In my opinion, relativistic quantum mechanics remains an open research field. There are still a few unresolved controversies. Students and readers which are visiting physicsforums.com are entitled to know that. I never tried to proclaim my ideas as given and commonly accepted facts. We all are learning here. And the best way to learn is through honest and corteous discussions. If you think that all issues raised in this thread have been clarified already, and the discussion should stop, then let's do that.

Regards.
Eugene.

 

[Fra]

There are still a few unresolved controversies. Students and readers which are visiting physicsforums.com are entitled to know that. I never tried to proclaim my ideas as given and commonly accepted facts. We all are learning here.

I agree with this. To try to tell students that things are simplier than they really are, and that their objections are without exception due to their ignorance is an indirect insult to their intelligence. Even though the latter may of course be a likely cause in many cases too.

I prefer a more honest attitude, to say that this may be one of our/my best description but if anyone/you thinks they have a better idea, don't let us/me discourage you.

/Fredrik

 

[jostpuur]

So we have two candidates for the probability density, $j^0$ and $|\phi|^2$. They both agree with the probability density $|\psi|^2$ of Shrodinger's equation in the non-relativistic limit, and they both remain conserved in the relativistic time evolution. So which one is correct?

Hans, you keep insisting that the $j^0$ is the correct one, because it is mainstream and it can be found in all books. Admittably that puts your opinion on stronger ground, but it should be in mainstream for a good reasons, and these reasons should be explainable.

We know that four vectors behave well in Lorentz transformations, but scalars behave quite well in Lorentz transformations too, so merely saying that probability must be a four vector so that it would transform like a four vector, isn't satisfying. Where are the true reasons?

It hasn't become clear to me, if there is a proof for the fact that $j^0$ is always nonnegative, or if there is a counterexample where it becomes negative. This matter needs more clarification.

I hope the discussion would start leaning more towards mathematics of the problem. Everyone's opinions are probably already clear to everyone.