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Symmetry and conserved probability current for wave functions

  1. Oct 3, 2007 #1
    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?
     
  2. jcsd
  3. Oct 3, 2007 #2
    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 [tex]\Psi(x,t)[/tex] is associated (via Noether's theorem) with the conserved probability density current?
     
  4. Oct 3, 2007 #3
    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.
     
  5. Oct 3, 2007 #4
    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.)
     
  6. Oct 3, 2007 #5
    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.
     
  7. Oct 3, 2007 #6
    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.)
     
    Last edited: Oct 3, 2007
  8. Oct 3, 2007 #7
    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.
     
  9. Oct 4, 2007 #8
    I'll continue from where I was left in the previous thread.

    The translation operator of the wave function is

    [tex]
    e^{u\cdot\nabla}.
    [/tex]

    It commutes with the Hamiltonian

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

    which implies

    [tex]
    [-i\hbar\nabla, H] = 0,
    [/tex]

    which implies, with the SE,

    [tex]
    D_t \langle\Psi |-i\hbar\nabla|\Psi\rangle = 0.
    [/tex]

    No we want to know why

    [tex]
    D_t \langle\Psi|\Psi\rangle = 0
    [/tex]

    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

    [tex]
    [1,H] = 0
    [/tex]

    in some sense, which instead would follow from the identity

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

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

    [tex]
    |\Psi\rangle\mapsto e^{i\theta}|\Psi\rangle.
    [/tex]

    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 [itex]\langle\Psi|\Psi\rangle[/itex] 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.
     
  10. Oct 4, 2007 #9
    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.
     
  11. Oct 4, 2007 #10

    nrqed

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    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.
     
  12. Oct 4, 2007 #11
    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.
     
  13. Oct 4, 2007 #12
    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.
     
  14. Oct 4, 2007 #13
    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.
     
  15. Oct 4, 2007 #14
    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.
     
  16. Oct 5, 2007 #15
    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 here, 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?
     
  17. Oct 5, 2007 #16

    Mentz114

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    pellman, the link to the google book search is unhelpful. I can't see eq. 10.41.
     
  18. Oct 5, 2007 #17
    The Lagrangian density for the one-particle Schrodinger wave-function is

    [tex]\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[/tex]

    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.)
     
    Last edited: Oct 5, 2007
  19. Oct 5, 2007 #18
    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 [itex] U_g^0 [/itex] 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 [itex] U_g [/itex] should be different from the non-interacting representation [itex] U_g^0 [/itex] constructed above. The fundamental question is how to construct [itex] U_g [/itex] 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.
     
    Last edited: Oct 6, 2007
  20. Oct 5, 2007 #19

    Mentz114

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    [tex]\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[/tex]

    Which is evidently invariant under

    [tex] \psi ' = e^{i\theta}\psi[/tex]
    [tex]\psi^{\dag}' = e^{-i\theta}\psi[/tex]

    and so is the probability current,


    [tex]\frac{\hbar^2}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)[/tex]

    [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.
     
    Last edited: Oct 5, 2007
  21. Oct 5, 2007 #20
    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.
     
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