Symmetry and conserved probability current for wave functions

[pellman]

Noether's theorem relates symmetries and conserved quantities, eg. if the Lagrangian is invariant under a spatial translation, you have have conservation of momentum. For continuous systems, the conserved quantities become conserved "currents".

Anyone know what symmetry is associated with the conservation of probability for the Schrodinger wave function?

 

[pellman]

This is the same question previously addressed in the Probability conservation and symmetry thread but never adequately answered.

To clarify, if we look at the Lagrangian density for the Schrodinger equation (the single particle case will do) what symmetry transformation on $$\Psi(x,t)$$ is associated (via Noether's theorem) with the conserved probability density current?

 

[meopemuk]

Noether's theorem relates symmetries and conserved quantities, eg. if the Lagrangian is invariant under a spatial translation, you have have conservation of momentum. For continuous systems, the conserved quantities become conserved "currents".

Anyone know what symmetry is associated with the conservation of probability for the Schrodinger wave function?

I am afraid in your question you are mixing two very different theories. Lagrangians and Noether's theorem are parts of the classical theory of fields. Conservation of probability is a feature of quantum mechanics. The conservation of probability in QM and its relationship to other conservation laws was worked out by Wigner in 1930's.

The first step in Wigner's theory is the famous Wigner's theorem which establishes that the conservation of probability implies that symmetries (such as inertial transformations of reference frames - translations, rotations, and boosts) must be represented by unitary (or antiunitary) operators in the Hilbert space of the system.

The second step is to realize that inertial transformations of reference frames must generate a unitary representation of the Poincare group in the Hilbert space. Then Hermitean operators of the total momentum, energy, and angular momentum are associated with generators of space translations, time translations, and rotations in this representation. The commutators between these operators follow from the structure of the Poincare Lie algebra. Operators commuting with the Hamiltonian correspond to conserved physical observables.

So, there is no any specific symmetry or observable associated with the conservation of probabilities. This conservation is reflected in the requirement of unitarity in quantum mechanics.

Eugene.

 

[cesiumfrog]

Is that basically saying "symmetry of the laws of QM under unitary transformations" is equivalent to conservation of probability? (That seems like what the OP is hoping for anyway, like e.g. symmetry of blah under translations is equivalent to conservation of momentum.)

 

[meopemuk]

Is that basically saying "symmetry of the laws of QM under unitary transformations" is equivalent to conservation of probability? (That seems like what the OP is hoping for anyway, like e.g. symmetry of blah under translations is equivalent to conservation of momentum.)

I would phrase it in a different way: Probabilities must be conserved with respect to inertial transformations (translations, rotations, etc.). This is a fundamental physical requirement. It then follows (via Wigner's theorem) that these transformations must be represented by unitary operators in the Hilbert space.

"Symmetry of the laws of QM under unitary transformations" is not a very meaningful statement, because a simultaneous unitary transformation of everything (both state vectors and operators of observables) in the Hilbert space is a trivial change of basis which doesn't affect physics in any way.

Eugene.

 

[cesiumfrog]

"Symmetry of the laws of QM under unitary transformations" is not a very meaningful statement, because a simultaneous unitary transformation of everything (both state vectors and operators of observables) in the Hilbert space is a trivial change of basis which doesn't affect physics in any way.

Um.. kindof like how a spatial translation is a trivial change of variables which doesn't affect physics in any way?

(Frankly, I don't have any opinion yet on whether your position is correct, I just find the argument so far to be particularly unpersuasive.)

 

[meopemuk]

Um.. kindof like how a spatial translation is a trivial change of variables which doesn't affect physics in any way?

No, this is not true. If two observers O and O' (O' is displaced relative to O) look at the same physical system, they would measure different values of the system's position. So, the unitary transformation of observables (from observables associated with observer O to observables associated with the displaced observer O') does make a difference for physical measurements.

It is a different matter if both observer and the state of the system (more precisely, the device which prepares the state of the system) are displaced together. Then we shouldn't expect any change in results (probabilities) of measurements. This follows from the principle of relativity. In the Hilbert space this simultaneous translation is described by an unitary transformation of both Hermitian operators of observables and state vectors. Of course, such a transformation leaves all expectation values unchanged.

Eugene.

 

[jostpuur]

I'll continue from where I was left in the previous thread.

The translation operator of the wave function is

$$e^{u\cdot\nabla}.$$

It commutes with the Hamiltonian

$$[e^{u\cdot\nabla}, H] = 0,$$

which implies

$$[-i\hbar\nabla, H] = 0,$$

which implies, with the SE,

$$D_t \langle\Psi |-i\hbar\nabla|\Psi\rangle = 0.$$

No we want to know why

$$D_t \langle\Psi|\Psi\rangle = 0$$

is true. I have difficulty believing that some derivation of this could be considered as the "reason" why this is true, but we can look for the analogy between the momentum conservation. This would follow from the identity

$$[1,H] = 0$$

in some sense, which instead would follow from the identity

$$[e^{i\theta}, H] = 0,$$

and this suggest that the related symmetry is the symmetry under the transformation

$$|\Psi\rangle\mapsto e^{i\theta}|\Psi\rangle.$$

I'm just throwing ideas here. Is this close to what meopemuk explained?

I'm not really sure what the "symmetries" are supposed to be in QM. In the classical context the L must be invariant under some transformations. Is it the $\langle\Psi|\Psi\rangle$ that is invariant now? Or I mean... I can see that it is invariant, but there is probably lot of invariant things, is this the relevant one?

In fact it would be interesting to see how the classical Noether's theorem gets derived out of some QM principles. Usually the Lagrange's function plays no role in QM.

 

[meopemuk]

In fact it would be interesting to see how the classical Noether's theorem gets derived out of some QM principles. Usually the Lagrange's function plays no role in QM.

The best place to learn about the connection between quantum mechanics and the theory of fields (classical and quantum) is in Weinberg's book "The quantum theory of fields", vol. 1. I often heard that this book is difficult to read. I have an opposite opinion. I was never able to comprehend the ad hoc logic of most other QFT textbooks, which usually begin from a theory of classical fields (with Lagrangians, Noether's theorem, etc.) and then introduce commutators between fields.

Weinberg's book is very different. Its logic is crystal clear. It begins with fundamental postulates of relativity and laws of quantum mechanics. Then Weinberg asks how one can construct the S-operator in multiparticle systems (which is the central quantity that connects theory with experiment) so that it is relativistically invariant, cluster separable, and allows for creation and annihilation of particles. He says that there is only one successful approach invented so far. This approach includes definition of quantum fields with certain commutation relations and transformation laws and then building the interacting Hamiltonian and boost operators as polynomials in these fields. In Weinberg's approach quantum fields are not "fundamental ingredients of nature", but some auxiliary formal mathematical objects that are found convenient for constructing the Hamiltonian and the S-operator.

I think this is the correct way to understand the connection between QM and field theory, Lagrangians, Noether's theorem, etc.

Eugene.

 

[nrqed]

The best place to learn about the connection between quantum mechanics and the theory of fields (classical and quantum) is in Weinberg's book "The quantum theory of fields", vol. 1. I often heard that this book is difficult to read. I have an opposite opinion. I was never able to comprehend the ad hoc logic of most other QFT textbooks, which usually begin from a theory of classical fields (with Lagrangians, Noether's theorem, etc.) and then introduce commutators between fields.

Weinberg's book is very different. Its logic is crystal clear. It begins with fundamental postulates of relativity and laws of quantum mechanics. Then Weinberg asks how one can construct the S-operator in multiparticle systems (which is the central quantity that connects theory with experiment) so that it is relativistically invariant, cluster separable, and allows for creation and annihilation of particles. He says that there is only one successful approach invented so far. This approach includes definition of quantum fields with certain commutation relations and transformation laws and then building the interacting Hamiltonian and boost operators as polynomials in these fields. In Weinberg's approach quantum fields are not "fundamental ingredients of nature", but some auxiliary formal mathematical objects that are found convenient for constructing the Hamiltonian and the S-operator.

I think this is the correct way to understand the connection between QM and field theory, Lagrangians, Noether's theorem, etc.

Eugene.

I agree 100%. I have alwasy disliked the "quantize a classical field" approach because to me it is a huge non sequitur. If I would teach QFT I could never get myself to tell that to students, it just does not make any sense to me. I have felt like I did not understand QFT for years because of this...I did the calculations but always felt that the very first step did not make any sense at all. And yet most books say that fields are the fundamental ingredients (even though those fields are never observed!)

To make things worse, in the same books it is explained that relativistic wave equations are flawed because they fail to take into account that the number of particles is not fixed in relativistic systems. So far so good. But then they say that the way to fix this is ...to quantize classical fields! What the?! Ofc ourse, after several pages of calculations, one sees the particle interpretation reemerge, but it feels like a HUGE non sequitur.

I have always felt " Well, if the problem is that the number of particles is not fixed, why not simply construct a Fock space and build the commutation relations and so on!? Why quantize classical field?


It's after 15 years of feeling that this is the way it should be done but never seeing it done anywhere that finally weinberg's book came out and it was a huge relief for me to see someone doing it the way that felt right to me.

To make things worse for students, there are many different ways QFT is introduced in books (I can think of 5 off hand!), all unrelated in their starting assumptions. It feels more like people trying to justify how to get an answer they know beforehand by using weird assumptions instead of attacking the problem in a logically consistent manner.
The only way that makes sense to me *at every step* is Weinberg's approach, with the fields coming out as a consequence of introducing varying number of particles instead of the usual other way around.

 

[jostpuur]

I have been considering getting the Weinberg's books. Sounds like I should. Unfortunately I'm a poor and busy student at the moment. But maybe later, once I get a job.

 

[meopemuk]

I agree 100%. I have alwasy disliked the "quantize a classical field" approach because to me it is a huge non sequitur. If I would teach QFT I could never get myself to tell that to students, it just does not make any sense to me. I have felt like I did not understand QFT for years because of this...I did the calculations but always felt that the very first step did not make any sense at all. And yet most books say that fields are the fundamental ingredients (even though those fields are never observed!)

To make things worse, in the same books it is explained that relativistic wave equations are flawed because they fail to take into account that the number of particles is not fixed in relativistic systems. So far so good. But then they say that the way to fix this is ...to quantize classical fields! What the?! Ofc ourse, after several pages of calculations, one sees the particle interpretation reemerge, but it feels like a HUGE non sequitur.

I have always felt " Well, if the problem is that the number of particles is not fixed, why not simply construct a Fock space and build the commutation relations and so on!? Why quantize classical field?


It's after 15 years of feeling that this is the way it should be done but never seeing it done anywhere that finally weinberg's book came out and it was a huge relief for me to see someone doing it the way that felt right to me.

To make things worse for students, there are many different ways QFT is introduced in books (I can think of 5 off hand!), all unrelated in their starting assumptions. It feels more like people trying to justify how to get an answer they know beforehand by using weird assumptions instead of attacking the problem in a logically consistent manner.
The only way that makes sense to me *at every step* is Weinberg's approach, with the fields coming out as a consequence of introducing varying number of particles instead of the usual other way around.

I agree with every word you wrote here. In my opinion, Weinberg's clear vision of the logic of QFT is his most significant contribution to physics. Even more significant than the electro-weak theory.

Eugene.

 

[meopemuk]

I have been considering getting the Weinberg's books. Sounds like I should. Unfortunately I'm a poor and busy student at the moment. But maybe later, once I get a job.

Hi jostpuur,

you should definitely get his vol. 1, read it slowly, and try to understand how and why his approach is different from numerous other QFT textbooks. I have nothing to say about his vol. 2. In my opinion, it is not much different from other similar books on the market. I think you can safely skip vol. 3 altogether.

Eugene.

 

[meopemuk]

There is a brief summary of Weinberg's views in the article

S. Weinberg, "What is quantum field theory, and what did we think it is?", http://www.arxiv.org/abs/hep-th/9702027

However, it is beyond me how he could write there

In its mature form, the idea of quantum field theory is that
quantum fields are the basic ingredients of the universe, and
particles are just bundles of energy and momentum of the fields.

In my opinion, this statement undermines the positive insight contained in his book.

Eugene.

 

[pellman]

Let me elaborate a bit further.

One can write down a field version of the Schrodinger theory. In the usual QM theory there is a distinct Hilbert space for a system of 1 particle, a space for 2 particles, etc. So you just use the union of all these spaces. (Nevermind differences of spin, charge, etc. Assuming identical particles here.) You wind up with an anihilation operator which obeys the single-particle Schrodinger equation (well, it's not quite that simple if you have an interaction potential.) That's your quantized field. The creation operator is the Hermitian of the anihilation operator. These two operators have equal-time commutators proportional delta functions, etc.

You don't gain anything by doing this since the states of different particle number remain orthogonal (no actual pair-creation or pair-annihilation). It's just regular old QM in field theory form.

And you can also then write down a Lagrangian for this field.

Or, heck, you can just write down a Lagrangian which yields the single-particle Schrodinger equation. Same thing.

This Lagrangian can be viewed http://books.google.com/books?id=XYc-YajwVuoC&pg=PA214&lpg=PA214&dq=schrodinger+lagrangian&source=web&ots=05qLzoAGm1&sig=wEtEz7nEsbvXroEwjwPtPGo3QvA#PPA216,M1, equation 10.41 Sorry, I haven't got this Tex thing down yet.

So the question is: is there a symmetry transformation of this Lagrangian which is associated with the conserved probability current, a la Noether's theorem?

Or, if you don't like this Schrodinger Lagrangian thing, I would be interested in the same question for the Dirac equation. The conserved current is discussed here. http://en.wikipedia.org/wiki/Dirac_equation#Adjoint_Equation_and_Dirac_Current Is there a symmetry of the Dirac Lagrangian density which is associated with this current?

 

[Mentz114]

pellman, the link to the google book search is unhelpful. I can't see eq. 10.41.

 

[pellman]

The Lagrangian density for the one-particle Schrodinger wave-function is

$$\mathcal{L}=-\frac{\hbar^2}{2m}\nabla\psi^{\dag}\cdot\nabla\psi+\frac{1}{2}i\hbar[\psi^{\dag}\stackrel{\cdot}{\psi}-\stackrel{\cdot}{\psi^{\dag}}\psi]-\psi^{\dag}V\psi$$

This is also the Lagrangian for the Schrodinger quantum field. (Though I think V has to be strictly an external potential--not an interaction--for this form.)

 

[meopemuk]

Let me elaborate a bit further.

One can write down a field version of the Schrodinger theory. In the usual QM theory there is a distinct Hilbert space for a system of 1 particle, a space for 2 particles, etc. So you just use the union of all these spaces. (Nevermind differences of spin, charge, etc. Assuming identical particles here.) You wind up with an anihilation operator which obeys the single-particle Schrodinger equation (well, it's not quite that simple if you have an interaction potential.) That's your quantized field. The creation operator is the Hermitian of the anihilation operator. These two operators have equal-time commutators proportional delta functions, etc.

You don't gain anything by doing this since the states of different particle number remain orthogonal (no actual pair-creation or pair-annihilation). It's just regular old QM in field theory form.

And you can also then write down a Lagrangian for this field.

Or, heck, you can just write down a Lagrangian which yields the single-particle Schrodinger equation. Same thing.

This Lagrangian can be viewed http://books.google.com/books?id=XYc-YajwVuoC&pg=PA214&lpg=PA214&dq=schrodinger+lagrangian&source=web&ots=05qLzoAGm1&sig=wEtEz7nEsbvXroEwjwPtPGo3QvA#PPA216,M1, equation 10.41 Sorry, I haven't got this Tex thing down yet.

So the question is: is there a symmetry transformation of this Lagrangian which is associated with the conserved probability current, a la Noether's theorem?

Or, if you don't like this Schrodinger Lagrangian thing, I would be interested in the same question for the Dirac equation. The conserved current is discussed here. http://en.wikipedia.org/wiki/Dirac_equation#Adjoint_Equation_and_Dirac_Current Is there a symmetry of the Dirac Lagrangian density which is associated with this current?

Hi pellman,

I understand well the reasons for your questions and your confusion. I was in the same state of confusion regarding QFT some 11-12 years ago before I read Weinberg's book and his earlier works, in particular

S. Weinberg, "The quantum theory of massless particles", in Lectures on Particles
and Field Theory, vol. 2, edited by S. Deser and K. W. Ford,
(Prentice-Hall, Englewood Cliffs, 1964)

which was the real eye-opener for me.

I strongly recommend you to find Weinberg's book and to study it. Then you would understand that questions you are asking don't make a lot of sense. Quantum fields have nothing to do with wave functions of particles. Equations satisfied by quantum fields (Klein-Gordon, Dirac, etc.) are not analogs of the Schroedinger equation. It doesn't make physical sense to write a Lagrangian from which the Schroedinger equation can be derived. Yes, you can do that mathematically, but this would contradict the fundamental logic of introducing quantum fields in the theory of multiparticle systems.

Quantum fields (and everything that go along with them - the Lagrangians, wave equations, etc.) are not useful in the theory of free non-interacting particles. Such a theory can be constructed in a simple non-controversial way without using the concept of fields at all. This can be done in a series of steps:

(1) Define Hilbert spaces of single particles as spaces of irreducible unitary representations of the Poincare group (see Wigner)

(2) Define multi-particle Hilbert spaces as tensor products of 1-particle spaces in (1).

(3) Define the Fock space as a direct sum of n-particle spaces, where n varies from 0 to infinity.

(4) Define particle creation and annihilation operators in the Fock space. Any (operator of) observable of physical interest now can be written as a polynomial in the particle creation and annihilation operators. Note that creation and annihilation operators can be written in any convenient representation. It is conventional to write them in the momentum-spin representation, but you can also write them in the (Newton-Wigner) position-spin representation.

(5) In the Fock space it is not difficult to define operators of particle observables (momentum, position, spin, etc.), corresponding eigenvectors, wave functions of multiparticle states (as expansion coefficients of the state vector in eigenvectors of particle observables). Moreover, it is easy to define the non-interacting unitary representation $U_g^0$ of the Poincare group which determines how operators of observables and wave functions transform with respect to time translations, space translations, rotations, and boosts. As usual, time translations of wave functions can be expressed as solutions of the Schroedinger equation.

The theory of free particles outlined above does not require any involvement of fields. According to Weinberg, quantum fields are necessary, if we want to consider inter-particle interactions. The theory of interacting particles is different from the above theory (1) - (5) only in one aspect: The unitary representation of the Poincare group $U_g$ should be different from the non-interacting representation $U_g^0$ constructed above. The fundamental question is how to construct $U_g$ so that all physical requirements (i.e., relativistic invariance, cluster separability, the possibility of particle creation and absorption, etc.) are satisfied? Weinberg shows how to do that by using certain formal linear combinations of particle creation and annihilation operators called "quantum fields".

In this approach the fundamental physical ingredients are particles and their interactions. Quantum fields are formal mathematical objects which do not have any physical interpretation and do not correspond to anything observed in nature. Their only role is to assist in derivation of particle interaction operators. The same formal status is assigned to all the machinery that goes with quantum fields: Lagrangians, wave equations, canonical quantization, gauge invariance, etc.

Surely, this is not how QFT is presented in most traditional textbooks. But I suggest you to stick to the Weinberg's interpretation. In my opinion, this is the only correct interpretation of QFT. The (legitimate) questions that you are asking don't have good answers in the traditional approach, but they simply don't make sense in the Weinberg's approach.

Eugene.

 

[Mentz114]

$$\mathcal{L}=-\frac{\hbar^2}{2m}\nabla\psi^{\dag}\cdot\nabla\psi +\frac{1}{2}i\hbar[\psi^{\dag}\stackrel{\cdot}{\psi}-\stackrel{\cdot}{\psi^{\dag}}\psi]-\psi^{\dag}V\psi$$

Which is evidently invariant under

$$\psi ' = e^{i\theta}\psi$$
$$\psi^{\dag}' = e^{-i\theta}\psi$$

and so is the probability current,


$$\frac{\hbar^2}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)$$

[edit]
Reading the posts of meopemuk I'm pursuaded that your question is not well framed, because it matters a great deal whether you mean a particle wave function or a field condition. The only time I've Noether's theorm applied it was in a classical contect.

The invariance of the probability ( charge) and probability current under SU(1) is self-evident.

 

[Anonym]

The only time I've Noether's theorm applied it was in a classical contect.

The purpose of the Exercise (10.5) is to demonstrate that the SE and its conjugate are Euler-Lagrange equations if the Lagrangian is given by (10.41). Then Noether's theorem applied as it was in a classical context with suitable changes of notations.

Where I am wrong?

Regards, Dany.

 

[pellman]

Thanks, meopemuk. I have further questions in response to your post. But I think I should start a new thread for that.

Let's forget about QFT for the moment.

Noether's theorem says that if the action is invariant under a certain transformation of the functions comprising the Lagrangian, then there is an associated conserved quantity. E.g., actions invariant under time translations represent systems with conservation of total energy. If the Lagrangian is a Lagrangian density--describing a continuous system--then the associated conserved quantity is a "current".

The Schrodinger single-particle wave function is such a continuous system. We can write down the Lagrangian density whose Euler-Lagrange equations yield the Schrodinger equation. And we know there is a conserved current

$$\frac{d\rho}{dt}+\nabla\cdot\boldmath j\unboldmath=0$$

where

$$\rho=|\psi|^2$$

and j is the current given above by Mentz114. So is the action invariant under a certain transformation of $$\psi$$ and/or x such that the associated conserved Noether current is the probability current?

It's just a math question really.

If the answer is yes, then I would expect to be a relatively commonly known sort of thing. If the answer is no, that's pretty interesting, I think.

Noether's theorem tells not only that a such a conserved current exists for a given symmetry but also how to construct it. Give me a symmetry and I'll give you a conserved current to go with it. But I don't think it works both ways. Given a conserved current, I don't know that there is a way to construct a symmetry of the Lagrangian to go with it.

 

[meopemuk]

It's just a math question really.

Exactly. I don't exclude the possibility that you can define some mathematical symmetry of your Lagrangian, whose corresponding conserved current is the "probability current". So what? In my opinion, it would be just a mathematical exercise without deep physical significance. This is just my personal opinion based on my (incomplete) understanding of quantum mechanics and field theory. I would be glad if you can prove me wrong.

Eugene.

 

[Anonym]

The Schrodinger single-particle wave function is such a continuous system. We can write down the Lagrangian density whose Euler-Lagrange equations yield the Schrodinger equation. And we know there is a conserved current

$$\frac{d\rho}{dt}+\nabla\cdot\boldmath j\unboldmath=0$$

where

$$\rho=|\psi|^2$$

and j is the current given above by Mentz114. So is the action invariant under a certain transformation of $$\psi$$ and/or x such that the associated conserved Noether current is the probability current?

If your statements are correct,, then we have small problem with Mentz114 post #19. Consider now the U(2) extension of that, known as Wienberg-Salam standard model... The mentioned above by Mentz114 U(1) symmetry remains exact and correspond to the conservation of the electric current (the Noether’s charge is now the generator of the transformations). Notice also that W-S so far has the complete experimental support.

The situation is as in the theater: the play is interesting but your place is occupied.

Regards, Dany.

 

[Mentz114]

pellman:

So is the action invariant under a certain transformation of $$\psi$$ and/or x such that the associated conserved Noether current is the probability current?

I have to say that applying Noethers theorem to the Lagrangian under question is dodgy.
But the answer is clearly 'yes'. You don't need Noether to see that 'charge' and 'current' are conserved under the gauge transformation, and so are the EOM from the Euler-Lagrange equations.

I did notice the following -

If we take a wave function which is a solution of the SE

$$\psi(x,t) = C e^{[\frac{i}{h}( Et - px )]}$$

and apply the gauge transformation it becomes
$$\psi(x,t) = C e^{[\frac{i}{h}( Et - px + \theta)]}$$

Now, can you see that theta is both a space and time translation. So the gauge transformation already covers the usual symmetries.

This is a very fast and dirty procedure but it shows the essential feature.

 

[Anonym]

This is a very fast and dirty procedure but it shows the essential feature.

Now, can you see that theta is both a space and time translation. So the gauge transformation already covers the usual symmetries.

Thus by the suitable choice of gauge you will eliminate the space and time altogether?

As you are all aware, now a days science-fiction writers use quantum physics to explain everything from near death experiences to global consciousness…
if someone here could find me a flippant (imagine a surly drunk physicist listening to dylan in a bar) quantum reply to this question that would be great: How many roads must a man walk down before you call him a man?

Imagine Alistair Maxwel dylan in a bar. Would you like to repeat your story?

Regards, Dany.

P.S. If you are ready to be slow and get idea how the gauge transformations are connected with the external symmetries, I suggest R.Utiyama, Phys. Rev., 101, 1597 (1956).

 

[pellman]

I have this figured now. The conserved current associated with the invariance of the Lagrangian under $$\psi\rightarrow e^{i\theta}\psi$$, where theta is a constant, is identical with the probability current.

Now we also know that for QFT allowing theta to vary with position is precisely the SU(1) symmetry whose gauge field is the EM potential. For the Klein-Gordon field, the associated conserved current represents conservation of charge, not probability.

I do think there is something deep here. It just needs some more thought.

Thanks for the responses. I am interested in further comments you may have.

 

[meopemuk]

I have this figured now. The conserved current associated with the invariance of the Lagrangian under $$\psi\rightarrow e^{i\theta}\psi$$, where theta is a constant, is identical with the probability current.
[...]
I do think there is something deep here. It just needs some more thought.

In my opinion, you just took a very simple thing (the conservation of probability, which is simply a part of its definition: the sum of probabilities of all alternative outcomes should be always equal to 1) and made it complicated and artificially "deep".

You wrote an ugly-looking Lagrangian, noticed that it is invariant with respect to $$\psi\rightarrow e^{i\theta}\psi$$, and then "derived" the law of conservation of probability. To me, this trick raises more questions than provides answers.

Eugene.

 

[Anonym]

I do think there is something deep here. It just needs some more thought.

For kids that want to know and understand physics. The discussions in this session were partially related with the presentation done in

N. A. Doughty, “Lagrangian Interaction, An Introduction to Relativistic Symmetry in Electrodynamics and Gravitation”, Addison-Wesley (1990).

I consider that textbook written as detective and outstandingly good and clear.

Regards, Dany.

 

[Micha]

I have this figured now. The conserved current associated with the invariance of the Lagrangian under $$\psi\rightarrow e^{i\theta}\psi$$, where theta is a constant, is identical with the probability current.

Now we also know that for QFT allowing theta to vary with position is precisely the SU(1) symmetry whose gauge field is the EM potential. For the Klein-Gordon field, the associated conserved current represents conservation of charge, not probability.

I do think there is something deep here. It just needs some more thought.

Thanks for the responses. I am interested in further comments you may have.

If a particle has U(1) symmetry, it is charged and its antiparticle has opposite charge. If your theory allows for creation of particle and antiparticle pairs, conservation of probability generalizes to conservation of charge. If you have a theory with no particle/antiparticle creation, you must have conservation of probability.

 

[pellman]

In my opinion, you just took a very simple thing (the conservation of probability, which is simply a part of its definition: the sum of probabilities of all alternative outcomes should be always equal to 1) and made it complicated and artificially "deep".

You wrote an ugly-looking Lagrangian, noticed that it is invariant with respect to $$\psi\rightarrow e^{i\theta}\psi$$, and then "derived" the law of conservation of probability. To me, this trick raises more questions than provides answers.

True, the conservation of probability does follow directly from the Schrodinger equation. We do not need to have a Lagrangian formulation to see that. But such a relation between the equations of motion and the Lagrangian is always the case, is it not? The equations of motion are sufficient. One can derive conservation of energy directly from Newton's laws without resorting to the Lagrangian or Hamiltonian formulations.

But there is something deep to the fact the time-translation invariance of the Lagrangian demands conservation of energy.

I would say the significance is something like this: if $$|\psi(x)|^2$$ is to be interpreted as a pdf, then it must satisfy a conservation law (so that it's normalization over all space is equal to 1 for all times). We also have the interpretation that the overall phase of $$\psi(x)$$ is physically unimportant. And here we see that the two things are intimately tied: if the overall phase of $$\psi(x)$$ did change the Lagrangian, then we would not have conservation of the quantity which we interpret as probability. The two things must go together.

 

[pellman]

If a particle has U(1) symmetry, it is charged and its antiparticle has opposite charge. If your theory allows for creation of particle and antiparticle pairs, conservation of probability generalizes to conservation of charge. If you have a theory with no particle/antiparticle creation, you must have conservation of probability.

I do not think that conservation of probability and conservation of charge are related in this way. The 0-compenent of the Dirac conserved current is positive definite. It does not represent charge since the Dirac equation describes both positrons and electrons. On the other, each of the four components of the Dirac spinor individually satisfy the Klein-Gordon equation and so also satisfy the Klein-Gordon conserved current, which does seem to be related to conservation of charge.

The Klein-Gordon pdf is what really interests me. I totally buy the fact that $$\phi\partial_{\mu}\phi^\dag-\phi^\dag\partial_{\mu}\phi$$ is the charge current. But then what is the pdf? Can't we still ask, what is the probability of observing a particle in such-a-such region? But you know what? The Klein-Gordon pdf is still an open question and people are still publishing papers on it.

 

[meopemuk]

I would say the significance is something like this: if $$|\psi(x)|^2$$ is to be interpreted as a pdf, then it must satisfy a conservation law (so that it's normalization over all space is equal to 1 for all times).

Yes, in quantum mechanics this is guaranteed by the fact that the Hamiltonian H is Hermitean and that the time evolution operator $U(t) = \exp(\frac{i}{\hbar} Ht)$ is unitary.

We also have the interpretation that the overall phase of $$\psi(x)$$ is physically unimportant.

Yes, in quantum mechanics this is guaranteed by the interpretation of states as rays of vectors in the Hilbert space, rather than individual vectors.

There is no need to introduce Lagrangian to "prove" these things. I still believe that Lagrangians are out of place in quantum mechanics. Yes, you can formally introduce some action, Lagrangian, etc. in quantum mechanics and "derive" the Schroedinger equation and "conservation of probabilities" from them. But this will not tell you anything new that you didn't know already from the basic laws of QM.

 

[meopemuk]

The Klein-Gordon pdf is what really interests me. I totally buy the fact that $$\phi\partial_{\mu}\phi^\dag-\phi^\dag\partial_{\mu}\phi$$ is the charge current. But then what is the pdf? Can't we still ask, what is the probability of observing a particle in such-a-such region? But you know what? The Klein-Gordon pdf is still an open question and people are still publishing papers on it.

You can find more discussions of the Klein-Gordon equation and the probabilistic interpretation of its solutions in a very long thread

https://www.physicsforums.com/showthread.php?t=175155

Eugene.

 

[Micha]

But there is something deep to the fact the time-translation invariance of the Lagrangian demands conservation of energy.

I would say the significance is something like this: if $$|\psi(x)|^2$$ is to be interpreted as a pdf, then it must satisfy a conservation law (so that it's normalization over all space is equal to 1 for all times). We also have the interpretation that the overall phase of $$\psi(x)$$ is physically unimportant. And here we see that the two things are intimately tied: if the overall phase of $$\psi(x)$$ did change the Lagrangian, then we would not have conservation of the quantity which we interpret as probability. The two things must go together.

I think, your idea is not bad. You should keep in mind though, that conservation of energy is a very general concept, which is problematic only in the context of general relativity.

Conservation of probability (which is also called unitarity), is also a very general concept. But the way, you present it here mathematically, as the norm of a single particle wave function, it is not very general. If you don't have a free theory, but allow for some interaction like electromagnetism, particles can be created out of energy.
So the fact, that the Schroedinger equation maintains the norm of the single particle wave function during time evolution, becomes a problem to fix for quantum field theory.