How observation leads to wavefunction collapse?

[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
<br /> \int d^3x |\phi|^2<br />
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&#039;}{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&#039;

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^*\phiNote 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&#039;}{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&#039;

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&#039;}{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}\ aRegards, 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.

 

[Hans de Vries]

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.

Hi Eugene,

I personally have no problem at all discussing these subjects as long as
it is constructive and the arguments are technical ones.

It would be highly appreciated that, if you want to discuss these issues,
you clearly identify which parts could be considered non-mainstream, this
as a simple courtesy to students who, for example, could loose points on
an exam as a result.

If you do so, then you relieve others of the task doing so, and you keep
the honor to yourself. Please don't feel forced to use counter offensive
political arguments as happened several times on this thread. The best
defense is being simply clear and open about the subject, and technical
in the discussion.


Regards, Hans

 

[jostpuur]

Hans, I was earlier in belief, that current j^0 must be used, because |\phi|^2 would not be conserved. I believe I have now got rid of that argument, so other equally convincing reasons to favor j^0 would be nice.

 

[Hans de Vries]

Hans, I was earlier in belief, that current j^0 must be used, because |\phi|^2 would not be conserved. I believe I have now got rid of that argument, so other equally convincing reasons to favor j^0 would be nice.

Jostpuur,

For the non-relativistic Schrödinger equation we have for the probability
density j_0 and the probability current density in the x_i direction j_i :

j_0\ =\ \Psi^*\Psi

j_i\ =\ \frac{i\hbar}{2m}\left[ \Psi^*\frac{\partial \Psi}{\partial x_i} - \frac{\partial \Psi^*}{\partial x_i}\Psi \right]

In the relativistic Klein Gordon equation all dimensions are on equal footing:

j_0\ =\ \frac{i\hbar}{2m}\left[ \psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi \right]

j_i\ =\ \frac{i\hbar}{2m}\left[ \psi^*\frac{\partial \psi}{\partial x_i} - \frac{\partial \psi^*}{\partial x_i}\psi \right]

The continuity relation says that the change in time of probability density
in a volume element dV is equal to the incoming minus the outgoing currents.

\partial_\mu j^\mu\ \ =\ \frac{\partial j_t}{\partial t} + \frac{\partial j_x}{\partial x} + \frac{\partial j_y}{\partial y} + \frac{\partial j_z}{\partial z} \ =\ 0

The continuity relation is valid in all reference frames in the case of the
Klein Gordon equation as well as in the rest frame of the Schrödinger
equation.

The continuity relation can be written as

\partial_\mu j^\mu\ \ =\ \psi^*M\psi-\psi M\psi* \ =\ 0

Where M is the Schroedinger equation and the Klein Gordon equation
operator respectively:M_{Sch} \ =\ \left[ \ i\frac{2m}{\hbar}\frac{\partial }{\partial t} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2}\ \right]

M_{KG}\ =\ \left[ \ \frac{\partial^2 }{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2} +\left( \frac{mc}{\hbar}\right)^2\ \right]
Regards, Hans

 

[jostpuur]

It is the equation

<br /> \frac{d}{dt}\int d^3x\; |\phi(t,x)|^2 = 0<br />

that is truly important. The continuity equation is only a one way to prove this, but I have now showed that this equation is true for positive frequency solutions of Klein-Gordon equation, even though the proof didn't use any kind of continuity equation.

 

[Hans de Vries]

It is the equation

<br /> \frac{d}{dt}\int d^3x\; |\phi(t,x)|^2 = 0<br />

that is truly important. The continuity equation is only a one way to prove this, but I have now showed that this equation is true for positive frequency solutions of Klein-Gordon equation, even though the proof didn't use any kind of continuity equation.

No, the problem is that this whole discussion about |\phi|^2 has hopelessly
confused you. It does not make any sense. It's plain wrong. The difference between the Schrödinger equation and the Klein Gordon
equation is in the definition of the wave function:

\psi\ =\ \Psi e^{-imt/\hbar}

Therefor, For stable solutions we get:

j_0\ =\ \frac{i\hbar}{2m}\left[ \psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi \right]\ =\ \Psi^*\Psi

...Regards, Hans

 

[gptejms]

Jostpuur,

For the non-relativistic Schrödinger equation we have for the probability
density j_0 and the probability current density in the x_i direction j_i :

j_0\ =\ \Psi^*\Psi

j_i\ =\ \frac{i\hbar}{2m}\left[ \Psi^*\frac{\partial \Psi}{\partial x_i} - \frac{\partial \Psi^*}{\partial x_i}\Psi \right]

In the relativistic Klein Gordon equation all dimensions are on equal footing:

j_0\ =\ \frac{i\hbar}{2m}\left[ \psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi \right]

j_i\ =\ \frac{i\hbar}{2m}\left[ \psi^*\frac{\partial \psi}{\partial x_i} - \frac{\partial \psi^*}{\partial x_i}\psi \right]

The continuity relation says that the change in time of probability density
in a volume element dV is equal to the incoming minus the outgoing currents.

\partial_\mu j^\mu\ \ =\ \frac{\partial j_t}{\partial t} + \frac{\partial j_x}{\partial x} + \frac{\partial j_y}{\partial y} + \frac{\partial j_z}{\partial z} \ =\ 0

The continuity relation is valid in all reference frames in the case of the
Klein Gordon equation as well as in the rest frame of the Schrödinger
equation.

Regards, Hans

Bumping into the discussion,isn't it wrong to interpret \psi(relativistic case) as a wavefunction and j_0 as a prob. density?I thought the original question was about transformation of (relativistic)wavefunction---satisfying Schrodinger equation(one of Eugene's posts).

 

[gptejms]

It is the equation

<br /> \frac{d}{dt}\int d^3x\; |\phi(t,x)|^2 = 0<br />

that is truly important. The continuity equation is only a one way to prove this, but I have now showed that this equation is true for positive frequency solutions of Klein-Gordon equation, even though the proof didn't use any kind of continuity equation.

Can you write down the proof--why does positive frequency solution of KG equation satisfy the above equation?

 

[gptejms]

No, the problem is that this whole discussion about |\phi|^2 has hopelessly
confused you. It does not make any sense. It's plain wrong.


The difference between the Schrödinger equation and the Klein Gordon
equation is in the definition of the wave function:

\psi\ =\ \Psi e^{-imt/\hbar}

Regards, Hans

This is true only for stationary states--what about more general states?

 

[jostpuur]

Can you write down the proof--why does positive frequency solution of KG equation satisfy the above equation?

I mentioned how the calculation can be carried out in my post 287, and posted some details below in post
290

This is true only for stationary states--what about more general states?

I thought that this is a non-relativistic approximation. In non-relativistic limit the rest mass energy dominates the oscillation. But I didn't go into details, I'm not sure.

 

[jostpuur]

The scientifical way to decide between j^0 and |\phi|^2 is of course nothing else than an experiment that tells which density is giving the correct probability density. I'm waiting eagerly to hear of such experiment.

 

[jostpuur]

How is probability density supposed to transform?

We cannot ask what is a probability for a particle to be at some specific location, but instead we can choose some small box of volume \Delta x^3, and ask what is probability for the particle to be in it. However, it doesn't make any sense to ask, what is the probability for a particle to be in this box in some boosted frame, because in such frame points of this box are not simultaneous anymore.

As result of this, I have difficulty in seeing how the probability density even should transform. Could somebody present some logical reasoning that would imply some transformation laws for the probability density?

 

[gptejms]

I mentioned how the calculation can be carried out in my post 287, and posted some details below in post
290

I've read the two posts you have referred to---interesting! You say in your post #287 '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.' Can you prove this statement i.e. show that the positive frequency solution of KG equation satisfies the relativistic Schrodinger equation?

 

[gptejms]

We cannot ask what is a probability for a particle to be at some specific location, but instead we can choose some small box of volume \Delta x^3, and ask what is probability for the particle to be in it. However, it doesn't make any sense to ask, what is the probability for a particle to be in this box in some boosted frame, because in such frame points of this box are not simultaneous anymore.

You can't ask for the 'probability for a particle to be in the box' at a given t(time in the original frame) but you can certainly ask for 'probability for a particle to be in the box' at a given t'(time in the boosted frame).

 

[Demystifier]

I'm waiting eagerly to hear Demystifier's explanations about how j^0 can get negative, indicating incompatibility with a probability interpretation.

Negative j^0 occurs when the wave function is a SUPERPOSITION of at least two different frequencies. For a simple example, see, e.g., Eq. (53) in
http://xxx.lanl.gov/abs/hep-th/0202204
and compare it with the nonrelativistic analog, Eq. (223).
The equations above are written in the language of QFT, but essentially the same results are obtained with first-quantized wave functions. Just take a superposition of two different frequencies and do it by yourself to convince yourself that the right-hand side of Eq. (53) will emerge.

 

[Demystifier]

So we want to calculate

<br /> \int d^3x\; \phi^*(t,\boldsymbol{x})\phi(t,\boldsymbol{x})<br />

Substituting the general solutions to this gives

<br /> =\int\frac{d^3x\;d^3p\;d^3y\;d^3p&#039;\;d^3y&#039;}{(2\pi)^6} \phi^*(0,\boldsymbol{y}) e^{i(E_{\boldsymbol{p}}t - \boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y}))} \phi(0,\boldsymbol{y}&#039;) e^{-i(E_{\boldsymbol{p}&#039;}t - \boldsymbol{p}&#039;\cdot(\boldsymbol{x}-\boldsymbol{y}&#039;))}<br />

This can be rearranged to be

<br /> =\int\frac{d^3x\;d^3p\;d^3y\;d^3p&#039;\;d^3y&#039;}{(2\pi)^6} \phi^*(0,\boldsymbol{y})\phi(0,\boldsymbol{y}&#039;) e^{i(E_{\boldsymbol{p}} - E_{\boldsymbol{p}&#039;})t} e^{i(\boldsymbol{p}\cdot\boldsymbol{y} - \boldsymbol{p}&#039;\cdot\boldsymbol{y&#039;})} e^{i(\boldsymbol{p}&#039;-\boldsymbol{p})\cdot\boldsymbol{x}}<br />

Integration of variable x can be carried out, and it gives \delta^{3}(\boldsymbol{p}&#039;-\boldsymbol{p}). Then you can integrate over the variable p', and this will remove energy terms.

You are right.
Very interesting result, congratulations!
Actually, your result does not depend on the equation of motion, all you need to know is that the equation is linear and that all energies are positive. Otherwise, the function E(p) may be arbitrary.
Still, there is a problem I see. It seems that you cannot generalize it to curved spacetime, because then the measure d^3x should be multiplied by an x-dependent quantity, so it is not clear that you will obtain delta-functions. On the other hand, the standard Klein-Gordon scalar product has a natural covariant generalization in curved spacetime. See, e.g., Eq. (59) in my "myths and facts" paper, or even better, Eqs. (8)-(9) in the reference mentioned in my previous post above.

 

[Hans de Vries]

Negative j^0 occurs when the wave function is a SUPERPOSITION of at least two different frequencies. For a simple example, see, e.g., Eq. (53) in
http://xxx.lanl.gov/abs/hep-th/0202204
and compare it with the nonrelativistic analog, Eq. (223).
The equations above are written in the language of QFT, but essentially the same results are obtained with first-quantized wave functions. Just take a superposition of two different frequencies and do it by yourself to convince yourself that the right-hand side of Eq. (53) will emerge.

Well, but a superposition of at least two different frequencies (in time)
does not satisfy the interaction free Klein Gordon or Dirac equation. Regards, Hans

 

[Demystifier]

Well, but a superposition of at least two different frequencies (in time)
does not satisfy the interaction free Klein Gordon or Dirac equation.

Yes it does. Both Klein-Gordon and Dirac equations are linear, so a superposition of solutions is also a solution. To avoid misunderstanding, I mean a superposition of wave functions that have different frequencies.

 

[Hans de Vries]

Yes it does. Both Klein-Gordon and Dirac equations are linear, so a superposition of solutions is also a solution.

They have to have equal m though.

To avoid misunderstanding, I mean a superposition of wave functions that have different frequencies.

Racing through the calculations I get something proportional to:

(\omega_1+\omega_2)\ (1 + \cos(\omega_1-\omega_2) )

which stays positive. It's almost equal to your equation (53). I'll have to
check this further.Regards, Hans

 

[Hans de Vries]

I'll have to check this further.

I still get the non-negative result:

j_0\ =\ \frac{i\hbar}{2m}\left[ \psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi \right]



In simplified notation with frequencies v and w:

(e^v + e^w)^*(ve^v+we^w) \ \ \ +\ \ \ (e^v + e^w)(ve^v+we^w)^*\ \ \ =

(e^{-v} + e^{-w})(ve^v+we^w) \ \ \ +\ \ \ (e^v + e^w)(ve^{-v}+we^{-w})\ \ \ =

v + w + ve^{v-w} + we^{-(v-w)}\ \ \ + \ \ \ v + w + ve^{-(v-w)} + we^{v-w}\ \ \ =

2(v+w) \ \ \ + \ 2v\cos(v-w)\ \ + \ \ 2w\cos(v-w)\ \ \ =

2(v+w) (1 + cos(v-w))



So the full end result would be:

j_0\ =\ \frac{(E_1+E_2)\hbar}{m}\ \left( 1\ +\ \cos((E_1-E_2)t/\hbar) \right)


Regards, Hans

 

[Demystifier]

Hans, you missed the correct normalization factor of wave functions, which also depends on frequency. See Eq. (41) and (42).

 

[Hans de Vries]

Hans, you missed the correct normalization factor of wave functions, which also depends on frequency. See Eq. (41) and (42).

This then gives your equation (53). Hmmm, this also means positive charge
density in a superposition of negative charge states...

I did encounter a problem with superpositions earlier. The point is that
the 4-current and the 4-momentum behave the same for single states
but differently for a superposition.

Take a superposition of two states with opposite speed:

e^{+ipx} + e^{-ipx}

Now the current density is zero but for the momentum one must take |p|
to get E right. This is somehow logical since everything is moving with
some speed v albeit in different directions, so one could expect E to
increase accordingly, even though the total momentum of the super-
position is zero.

Assuming that a wavefunction transforms \phi_\mu(x)\rightarrow \phi(\Lambda x)[/itex] then j_\muhas<br /> to transform like a 4-vector <u>momentum</u> to correct for the Lorentz<br /> contraction.<br /> <br /> Now, the 4-vector current and the 4-vector momentum do not behave <br /> the same anymore for a superposition of states.<br /> <br /> Given the fact that the different states all have different speeds makes me<br /> doubt that we can even Lorentz transform a superposition as a whole.<br /> If the states have speeds +v and -v and one boost by v then one state<br /> gets v=0 and the other 2v (for small v)<br /> <br /> Trying to figure things out now along this direction...Regards, Hans

 

[meopemuk]

Still, there is a problem I see. It seems that you cannot generalize it to curved spacetime, because then the measure d^3x should be multiplied by an x-dependent quantity, so it is not clear that you will obtain delta-functions. On the other hand, the standard Klein-Gordon scalar product has a natural covariant generalization in curved spacetime.

This doesn't seem to be a valid argument. It presumes that we know how the quantum theory of gravity will look like (curved spacetime and all that). But, as you well know, there is no consensus about that.

Eugene.

 

[Demystifier]

This then gives your equation (53). Hmmm, this also means positive charge
density in a superposition of negative charge states...

I did encounter a problem with superpositions earlier. The point is that
the 4-current and the 4-momentum behave the same for single states
but differently for a superposition.

Take a superposition of two states with opposite speed:

e^{+ipx} + e^{-ipx}

Now the current density is zero but for the momentum one must take |p|
to get E right. This is somehow logical since everything is moving with
some speed v albeit in different directions, so one could expect E to
increase accordingly, even though the total momentum of the super-
position is zero.

Assuming that a wavefunction transforms \phi_\mu(x)\rightarrow \phi(\Lambda x)[/itex] then j_\muhas<br /> to transform like a 4-vector <u>momentum</u> to correct for the Lorentz<br /> contraction.<br /> <br /> Now, the 4-vector current and the 4-vector momentum do not behave <br /> the same anymore for a superposition of states.<br /> <br /> Given the fact that the different states all have different speeds makes me<br /> doubt that we can even Lorentz transform a superposition as a whole.<br /> If the states have speeds +v and -v and one boost by v then one state<br /> gets v=0 and the other 2v (for small v)<br /> <br /> Trying to figure things out now along this direction...<br />

<br /> Concerning your first remark, note that the particle current can be introduced even for neutral particles (see again my paper). Therefore, the negative j_0 should NOT be interpreted as a negative charge density.<br /> <br /> Concerning your puzzle about current and velocity, recall that velocity is UNCERTAIN in a superposition. Therefore, I am not sure that it makes sense to talk about the Lorentz transformation of velocity. Note also that velocity is not uncertain in the Bohmian interpretation, in which the velocity is identified with the current at the point at which the pointlike particle is.

 

[jostpuur]

I've read the two posts you have referred to---interesting! You say in your post #287 '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.' Can you prove this statement i.e. show that the positive frequency solution of KG equation satisfies the relativistic Schrodinger equation?

What you can show quite easily is that the solutions of the relativistic Shrodinger equation are also solution of the Klein-Gordon equation, with the positive frequency. First show that

<br /> \sqrt{-\nabla^2 + m^2}\big(\sqrt{-\nabla^2 + m^2}\psi(x)\big) = \big(-\nabla^2 + m^2\big)\psi(x)<br />

This is easy to believe, but you can also calculate it starting with an explicit position representation definition

<br /> \sqrt{-\nabla^2 + m^2}\psi(x) := \int\frac{d^3x&#039;\;d^3p}{(2\pi)^3}\psi(x&#039;)\sqrt{|p|^2+m^2}e^{ip\cdot(x-x&#039;)}<br />

Take a time derivative of the both sides of the Shrodinger equation, substitute the original Shrodinger equation on the right hand side, and you get the Klein-Gordon equation. Checking that the solution is positive frequency, and that it exists in the first place, is not difficult either because we already know how to write down solution of Klein-Gordon equation as linear combinations of the plane waves, and

<br /> \psi(t,x) = \int\frac{d^3p\;d^3x&#039;}{(2\pi)^3} \psi(0,x&#039;)e^{-i(E_p t + p\cdot(x&#039;-x))}<br />

is the only obvious solution candidate. If you substitute it into the relativistic Shrodinger equation, you see it solves it.

hmh... I just realized that the first part of this response is redundant. Well I have now being changing this response so intensely, that I'll let it be like this.

 

[meopemuk]

What you can show quite easily is that the solutions of the relativistic Shrodinger equation are also solution of the Klein-Gordon equation, with the positive frequency.

Yes, your derivation is correct. However, why did you stop at taking the first time derivative of the Schroedinger equation? Why didn't you take 2nd derivative, 3rd derivative, etc? By doing that, you would obtain an infinite number of equations that are 3rd order in t, 4th order in t, etc. Can you say that all these equations are equivalent to the original Schroedinger equation? Apparently, not. The reason is that all these equations do not allow you to predict wavefunction at time t>0 if you know the wavefunction at t=0. Only the original Schroedinger equation can do that, because it is of the first-order in t. Is this an important distinction? I think, yes. Because in quantum mechanics the wavefunction (not wavefunction + its time derivatives, but the wavefunction alone) is supposed to provide a complete description of the state. Therefore, the wavefunction alone at t=0 must fully determine the time evolution of the state at later times. Therefore, only the first-order Schroedinger equation gives a complete picture of time evolution.

Yes, you can also use the 2nd-order Klein-Gordon equation (or even higher order equations) to describe the time evolution of wavefunctions, but then you need to supplement it with information about time derivatives of the wavefunction at t=0. So, the Klein-Gordon equation doesn't tell the full story.

Eugene.

 

[jostpuur]

Yes, your derivation is correct. However, why did you stop at taking the first time derivative of the Schroedinger equation? Why didn't you take 2nd derivative, 3rd derivative, etc? By doing that, you would obtain an infinite number of equations that are 3rd order in t, 4th order in t, etc. Can you say that all these equations are equivalent to the original Schroedinger equation? Apparently, not. The reason is that all these equations do not allow you to predict wavefunction at time t>0 if you know the wavefunction at t=0. Only the original Schroedinger equation can do that, because it is of the first-order in t. Is this an important distinction? I think, yes. Because in quantum mechanics the wavefunction (not wavefunction + its time derivatives, but the wavefunction alone) is supposed to provide a complete description of the state. Therefore, the wavefunction alone at t=0 must fully determine the time evolution of the state at later times. Therefore, only the first-order Schroedinger equation gives a complete picture of time evolution.

Yes, you can also use the 2nd-order Klein-Gordon equation (or even higher order equations) to describe the time evolution of wavefunctions, but then you need to supplement it with information about time derivatives of the wavefunction at t=0. So, the Klein-Gordon equation doesn't tell the full story.

Eugene.

I mentioned that the Klein-Gordon equation must be equipped with the additional information, that only positive frequency solutions are considered. Then it defines unique time evolution.

I admit that the Shrodinger equation seems to be more fundamental source of the time evolution, but its connection to the Klein-Gordon equation should not be ignored. Noting that the solutions of Shrodinger equation are also solutions of Klein-Gordon equation makes it easier to understand what kind of solutions they are. After this you also understand better where the Klein-Gordon equation is coming from.

 

[Johan de Vries]

Hi all

I know I raised a similar question in the thread "Wave particle duality", but it is already so full of many other questions, that I'd not be able to discuss this topic fully there.

So, in the double slit experiment, if a photon observes an electron, the interference pattern vanishes. Why is this so? What does a photon do to an electron? Also, can anybody explain to me as to how a single electron creates an interference pattern in reality? I am completely at sea as far as understanding this phenomenon is concerned. I know that in theory we have wavefunctions, but how can all the paths that can be followed by the electron, consist of one in which it passes through both the slits?

thanks
Mr Virtual

(I didn't see my favorite way to explain this to beginning students...)

If you do a double slit experiment with electrons and you use photons to observe through which slit they go, then you don't get an interference pattern. If you don't observe the electrons then the wavefunction is like:

|psi_1> + |psi_2>

were |psi_i> is the contribution to the wavefunction from slit i. But if the electron interacts with photons then you get a superposition like:

|psi_1>| ph_1> + |psi_2>| ph_2>

where |ph_i> is the wave function for the photon in case the electron moves through slit i. Suppose we don't observe the photon, and we want to find the probability that the electron will end up at some postion on the screen behind the two slits. The probability of finding a system in som state is given by the absolute value squared of the inner product with that state. So, in case of no interaction wit photons you get for the probability:

|<x|psi_1> + <x|psi_2>|^2 =

|psi_1(x) + psi_2(x)|^2

And this thus contans interference terms. In the case of interaction with photons you have to take the inner product with the state corresponding to the photon being in state |x> and the photon being in some state |ph>, take the absolute value square and sum over a complete set of photon states |ph>:

sum over |ph> of:

|<x|psi_1><ph| ph_1> + <x|psi_2><ph| ph_2>|^2


Now let's look at the interference term:

Sum over |ph> of:

2 Re[psi_1(x)psi_2-star(x) <ph_2| ph><ph| ph_1> ] =

2 Re[psi_1(x)psi_2-star(x) <ph_2|ph_1> ]

So, the interference term is proportional to the overlap of the photon wave functions |ph_1> and |ph_2>. If there is no overlap then the photon "knows for sure" through which slit the electron went

 

[gptejms]

Take a time derivative of the both sides of the Shrodinger equation, substitute the original Shrodinger equation on the right hand side, and you get the Klein-Gordon equation. Checking that the solution is positive frequency, and that it exists in the first place, is not difficult either because we already know how to write down solution of Klein-Gordon equation as linear combinations of the plane waves, and

<br /> \psi(t,x) = \int\frac{d^3p\;d^3x&#039;}{(2\pi)^3} \psi(0,x&#039;)e^{-i(E_p t + p\cdot(x&#039;-x))}<br />

is the only obvious solution candidate. If you substitute it into the relativistic Shrodinger equation, you see it solves it.

What I understand of positive and negative frequency solutions is this:

If
f(t)=\int_{-\infty}^{\infty}f(\omega)e^{-\iota \omega t}d\omega,
then if f(t) is real it can be written as

f(t)=\int_{0}^{\infty}f^{*}(\omega)e^{\iota \omega t}d\omega +<br /> \int_{0}^{\infty}f(\omega)e^{-\iota \omega t}d\omega,

where f^{*}(\omega)=f(-\omega)
The first term on RHS may be called the negative frequency term and the second the positive frequency term.Now in your posts you don't seem to be doing this.Your range of integration seems to be from -infinity to infinity always--there would be a problem if you were to show your results with the above definition.I don't know what limits the (authors of) books have in mind when they talk of +ve and -ve freq. terms(but I guess they take your definition).

 

[jostpuur]

What I understand of positive and negative frequency solutions is this:

If
f(t)=\int_{-\infty}^{\infty}f(\omega)e^{-\iota \omega t}d\omega,
then if f(t) is real it can be written as

f(t)=\int_{0}^{\infty}f^{*}(\omega)e^{\iota \omega t}d\omega +<br /> \int_{0}^{\infty}f(\omega)e^{-\iota \omega t}d\omega,

where f^{*}(\omega)=f(-\omega)
The first term on RHS may be called the negative frequency term and the second the positive frequency term.Now in your posts you don't seem to be doing this.Your range of integration seems to be from -infinity to infinity always--there would be a problem if you were to show your results with the above definition.I don't know what limits the (authors of) books have in mind when they talk of +ve and -ve freq. terms(but I guess they take your definition).

The plane wave solutions of Klein-Gordon equations are

e^{-i(E_p t-p\cdot x)} (positive frequency)

and

e^{i(E_p t + p\cdot x)} (negative frequency)

(Actually I'm not sure about the sign convention for the term p\cdot x here. Perhaps there is a minus sign. It isn't important anyway, it leads only into a redefinition of \phi^-_p)

An arbitrary wave packet may be written as superposition of these like this

<br /> \phi(t,x) = \int\frac{d^3p}{(2\pi)^3} \Big( \phi^+_p e^{-iE_p t} + \phi^-_p e^{iE_p t}\Big) e^{ip\cdot x}<br />

Where the p is integrated over the space \mathbb{R}^3, and \phi^+_p and \phi^-_p are functions of p (I prefer this notation for Fourier transformations). If we are only interested in the positive frequency solutions, then the wave packet is of form

<br /> \phi(t,x) = \int\frac{d^3p}{(2\pi)^3} \phi^+_p e^{-i(E_p t - p\cdot x)}<br />

Notice that E_p=\sqrt{|p|^2+m^2} is getting only positive values, when p is integrated over the three dimensional real space. At time t=0 this becomes a Fourier transform of the \phi^+_p, so for a desired initial wave packet \phi(0,x) we can substitute an inverse transform

<br /> \phi^+_p = \int d^3x&#039;\; \phi(0,x&#039;) e^{-ip\cdot x&#039;}<br />

and then we get to the solution I wrote earlier.

Trying to write wave packets by integrating over the energy (or frequency) doesn't look very practical, because energy doesn't define the momentum uniquely.

 

[gptejms]

The plane wave solutions of Klein-Gordon equations are

e^{-i(E_p t-p\cdot x)} (positive frequency)

and

e^{i(E_p t + p\cdot x)} (negative frequency)

(Actually I'm not sure about the sign convention for the term p\cdot x here. Perhaps there is a minus sign. It isn't important anyway, it leads only into a redefinition of \phi^-_p)

There is a minus sign there--otherwise what is the negative frequency here--p is also a (spatial) frequency(though you can correlate the negative (or positive)frequency with the sign of E_p also)

Trying to write wave packets by integrating over the energy (or frequency) doesn't look very practical, because energy doesn't define the momentum uniquely.

I took f to be a function of t and integrated over frequency just to illustrate the point--it could well be a function of x(in which case you'd be dealing with k,the spatial frequency or wavenumber)

 

[ghostofthecat]

a strange theory

Hi folks,

I am new to the study of quantum physics, so I humbly state that I am here to learn mostly, however, after reading much about the double-slit experiment and wavefunction, I would like to test a theory, if the wavefunction collapses under observation, could it mean that by observing the phenomenon, we impose time and space perception to it and that is what causes it to collapse? Could this also be a clue to time-travel. I have read other theories about human perception changing time and space, but I wonder if it fits into this experiment. Sorry if it sounds hokey but its just an idea I had and I would love to hear any feedback.

Thanks,
the ghost of schrodingers cat

 

[Mr Virtual]

As far as I know, physicsists still don't know the real reason for this collapse. This is what I have gathered from all the discussion so far.
As far as your own theory goes, someone well versed in QM maths/theory may answer your question.

regards
Mr V

 

[Anonym]

As far as I know, physicsists still don't know the real reason for this collapse. This is what I have gathered from all the discussion so far.
As far as your own theory goes, someone well versed in QM maths/theory may answer your question.

The real reason for the collapse of the wave packet was explained originally by A. Einstein in 1927: our world is 4-dim space-time continuum with the Minkowski metric. However, there are people that deny the validity of special relativity and QM. I leave for your judgment whether they may be considered physicists.

The only way to understand that is:

Mr.Virtual -- If you take a 1st year grad level QM course, or equivalent thereof, you'll find most of your questions answered -- in fact, these questions have been around for almost a century, and very sophisticated answers and arguments abound in the literature.

Regards, Dany.

 

[Mike2]

Is it true that the wave function describes propagation in one direction in time? But if it does describe propagation in time, then it can not give information of both initial and final states at the same time, since it propagated from one to the other in time. So there's no information of the initial state to enable a calculation of probabilities from initial to final states; the final state could have come from many different initial states. In order to determine the probability of going from initial and final states, we have to have the reverse propagation from final to initial state. Then we know both intial and final states enabling a calculation of the probability from initial to final state. Thus the wave function is multiplied by its complex conjugate to cancel out the time dependencies and get information of both initial and final states at the same instant in order to get simultaneous knowledge of both events required to "know" at some instant the probability of going from one to the other. Does this all sound right?

 

[orange juice]

Hi all

I know I raised a similar question in the thread "Wave particle duality", but it is already so full of many other questions, that I'd not be able to discuss this topic fully there.

So, in the double slit experiment, if a photon observes an electron, the interference pattern vanishes. Why is this so? What does a photon do to an electron? Also, can anybody explain to me as to how a single electron creates an interference pattern in reality? I am completely at sea as far as understanding this phenomenon is concerned. I know that in theory we have wavefunctions, but how can all the paths that can be followed by the electron, consist of one in which it passes through both the slits?

thanks
Mr Virtual

I think The Feynman Lecture on Physics (Vol.3), section 3-2 "the two-slit interference pattern" can answer some of your puzzles, and as Feynman said there, "although we don't have the correct mathematical formula for all the factors that go into this calculation, you will see the spirit of it in the following discussion."