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.