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Gauge dependent probabilities

  1. Nov 1, 2011 #1

    Jano L.

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    I was trying to understand the standard calculation of Einstein's A,B coefficients in quantum theory textbooks and I came across the following difficulty. In the calculation of transition probabilities, total wavefunction is expanded into eigenstates of time-independent [itex]H_0[/itex] and the expansion coefficients are interpreted as giving the frequencies of occurence of the corresponding states. For example, for the hydrogen atom Hamiltonian [itex] \hat H_0 = -\frac{\hbar^2}{2m}\Delta + q_e\varphi(r) [/itex]

    [tex]
    \psi(\mathbf r, t) = \sum_k c_k(t) \phi_k(\mathbf r)
    [/tex]

    where

    [tex]
    \hat H_0 \phi_k (\mathbf r) = \epsilon_k \phi_k(\mathbf r)
    [/tex]

    Plugging the expansion into the time dependent Schroedinger's equation, we can calculate all [itex]c_k[/itex]s. The accepted interpretation of these numbers is that [itex] c_k^* c_k[/itex] is the probability that randomly chosen atom will be in the [itex]k[/itex]-th stationary state.

    The problem is, that these numbers are gauge-dependent. How can a probability depend on the choice of gauge?

    Proof:

    In the initial gauge [itex] \varphi = Kq_p/r, \mathbf A = 0 [/itex]:

    [itex] c_k (t) = \langle \phi_k | \psi(t) \rangle = \int \phi_k(\mathbf r) \psi(\mathbf r, t) dV [/itex]

    In another gauge [itex] \varphi' = \varphi - \partial_t \chi, \mathbf A ' = \mathbf A + \nabla \chi, \psi' = e ^{-i\frac{q}{\hbar}\chi} \psi[/itex]:

    [itex] c_k' (t) = \langle \phi_k | \psi(t)'\rangle = \int \phi_k(\mathbf r) e^{-i\frac{q}{\hbar} \chi(\mathbf r, t)}\psi(\mathbf r, t) dV [/itex]

    which can be made equal to any number by suitable choice of [itex]\chi [/itex], for example, time independent [itex] \chi [/itex] rapidly oscillating in spatial coordinates will make it close to zero without change in the field strength (provided it has zero curl, which is not hard to fulfill).

    One may object that in the second gauge one should calculate coefficients using new functions for the stationary states, somehow like this:

    [itex] c_k' = \langle \phi_k' (\mathbf r, t) | \psi(t)' \rangle [/itex]

    where [itex] \phi_k'(\mathbf r, t)[/itex] is a stationary state in the new gauge, but how to find such function? Clearly [itex] \phi_k'(t) = e^{-i\frac{q}{\hbar}\chi} \phi_k [/itex] won't work, because then we'll get
    [tex]
    \hat H'(t) \phi_k'(t) = [ \epsilon_k - q_e\partial _t \chi (\mathbf r, t)] \phi_k'(t)
    [/tex]
    with energy eigenvalue dependent on the spatial coordinates as well.

    Or is there some other gauge-invariant way to calculate the A, B coefficients?
     
  2. jcsd
  3. Nov 4, 2011 #2

    Jano L.

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    I know it is a very difficult question. But there must be some ideas...?
     
  4. Nov 5, 2011 #3

    strangerep

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    I just didn't have very much spare time lately. The "difficult" part is in de-constructing what you're calculations to discover the mistake...

    Something looks wrong here - you've written the new Hamiltonian as time-dependent, whereas a physical Hamiltonian for a conservative system should be time-independent.

    Transformations like
    [tex]
    \phi' ~=~ \phi + z
    [/tex]
    can usually be implemented via a exponentiated operator like this:
    [tex]
    U(z) ~:=~ e^{z\phi - \bar{z} \phi^\dagger}
    [/tex]
    (where I might have a sign wrong). This is standard in the theory of ordinary coherent states. In your case, I guess [itex]z[/itex] must be time-dependent.

    I haven't worked this out all the way (not enough spare time), so I'm not sure it answers your question. Maybe you can progress the calculations?
     
  5. Nov 5, 2011 #4

    tom.stoer

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    Using a time-dependent gauge function transforms your problem into a non-stationary problem with time-dependent Hamiltonian H(t) due to a time-dependent el.-mag. field. But of course physically (i.e. in all observables, spectra etc.) no time dependency must be present b/c this is 'pure gauge' and therefore articial.

    I think you have to start with the full Hamiltonian of a quantum particle in an el.-mag. field. In the special case of the Coulomb potential the A-terms will drop out. Of course in the new gauge these terms are present and you will get new, gauge dependent solutions phi. The transformation of the Hamiltonian and of the eigenfunctions w.r.t. gauge transformations are 'standard textbook'.

    http://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html

    After having done that you should be able to see how to proceed with the coefficients c(t).
     
  6. Nov 15, 2011 #5

    Jano L.

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    "Something looks wrong here - you've written the new Hamiltonian as time-dependent, whereas a physical Hamiltonian for a conservative system should be time-independent."

    I think that there is no such thing as "physical Hamiltonian". The usual Hamiltonian is time-independent and is easy to interpret as kinetic+potential energy, but this is not necessary. It is easily shown that for the calculation of [itex]\psi(t)[/itex] of the electron in hydrogen atom we can use the Hamiltonian

    [itex]
    \hat H_0 ' = \frac{(-i\hbar \nabla - q \mathbf A)^2}{2m},
    [/itex]

    where

    [itex]
    \mathbf A(\mathbf r, t) = k\frac{\mathbf r}{r^3} t.
    [/itex]

    This is also easy to interpret, as the kinetic energy of the electron.

    In some situations it is not possible to have time-independent Hamiltonian at all. Consider an electron in a slowly varying magnetostatic field - the Hamiltonian will necessarily contain [itex]\mathbf A[/itex] and this would depend on time. (Born's theory of adiabatic perturbations).

    The important thing is, that neither in the Hamiltonian equations nor in Schrodinger's equation there is any preference for a particular gauge. These equations are gauge-invariant and do not introduce any special basis of eigenfunctions. In Schrodinger's equation, the important gauge-independent quantity is [itex] |\psi(x,y,z,t)|^2[/itex].

    It is only the need for some mathematical representation of the notion of stationary states that leads to search for the eigenfunctions. But these depend on the [itex]H_0[/itex] chosen (first choice).

    Furthermore, for chosen [itex]H_0[/itex], there is still ambiguity as to the choice of the gauge for the external field. Different gauges give different scalar products with the chosen eigenfunctions, but one has to be chosen. (second choice).

    So there are two questions:

    1) why do we choose [itex]H_0[/itex] to find eigenfunctions and not, say, [itex]H_0'[/itex] ?
    2) if we have chosen [itex]H_0[/itex], which form of the potentials [itex]\mathbf A, \phi[/itex] for the external field should we use to calculate [itex] \psi(t)[/itex]?

    I do not see any natural way to make these two choices. Fortunately, no such choice is required to interpret [itex]|\psi|^2[/itex]. I think all interpretations should be based on this quantity, not on expansion coefficients.

    It can be said that every consistent set of choices 1,2 is equally right, but then we are clearly calculating different probabilities for different choices. Which one is relevant for the calculation of Einstein's coefficients?
     
    Last edited: Nov 15, 2011
  7. Nov 15, 2011 #6

    Ken G

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    It sounds to me that the answer to your question lies in certain hidden assumptions that underlie the whole scheme you applied, wherein the coefficients are interpreted as "the probability that randomly chosen atom will be in the k-th stationary state." What requirements must be fulfilled for that interpretation to hold true, and how might those requirements be violated in certain gauges? I don't know the answer off hand, but I suspect that must be the place where the answer lives. We can certainly agree that the transition rates must not depend on the gauge chosen, so if you really do get different answers in different gauges, then one of the gauges was not applied consistently with the assumptions used to interpret the transition rates.
     
  8. Nov 15, 2011 #7

    strangerep

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    Jano,

    I'm unable to give a helpful answer, given the way you've expressed the question. You make too many sweeping statements without giving specific textbook references. If I knew precisely what you were reading, perhaps I could untangle what assumptions are involved.

    I suggest Ballentine ch11 as a basis for further discussion, if you can obtain a copy. (If not, maybe Google books or Amazon will let you read enough of pp312-313 to form a basis for further discussion.)

    Ballentine does indeed start from that point (see p312).
     
  9. Nov 16, 2011 #8
    Jano,

    I was interested in similar questions a long time ago. I didn't find a good answer then, but I remember that there were lots of journal papers studying this issue, but no consensus. I don't have the references now. You can try your own citation search, starting, for example, from this paper and references therein:

    Chen, C. Y., Transition probability and preferential gauge, quant-ph/9907058

    Regards.
    Eugene.
     
  10. Nov 16, 2011 #9
  11. Nov 16, 2011 #10
    Something is clearly not OK here, you cannot have space dependent eigenvalues. Inspired by http://iopscience.iop.org/0305-4470/20/3/023", I did some calculations. I do not have the general answer, but here are some hints:

    Transformation [itex]\chi(r,t)=\chi(r)[/itex]:
    Here, it is pretty simple. I start with time-dependent Schroedinger equation [tex]i\hbar \partial_t\phi_n=\hat{H}_0 \phi_n\;,[/tex]which is gauge independent if we choose gauge-independent [itex]\hat{H}_0[/itex], so [tex]i\hbar \partial_t\phi'_n=\hat{H}'_0 \phi'_n\;.[/tex]Eigenvalues are defined by stationaty Schroedinger equation[tex]\hat{H}_0\bar{\phi}_n=\epsilon_n\bar{\phi}_n\;,[/tex] which is obtained after ansatz for stationary states [tex]\phi_n=\exp(i\epsilon_n t/\hbar)\bar{\phi}_n\;.[/tex]If we have time-independent [itex]\chi(r)[/itex], we didn't spoil the calibration invariance of the stationary Schroedinger's Eq., because it differs from time-dependent Schroedinger equation only by factor [itex]\exp(i\epsilon_n/\hbar t)[/itex], which contains no spatial-dependence or [itex]\chi[/itex]-dependence, therefore (see also the article above):
    [tex]\hat{H}'_0\bar{\phi}'_n=\epsilon_n\bar{\phi}'_n\;,[/tex] where [itex]\bar{\phi}'_n=\exp(i \frac{e}{\hbar}\chi(r))\bar{\phi}_n[/itex]

    Transformation [itex]\chi(r,t)=\chi(t)[/itex]:
    Here, we have not a problem with the spatial part of the transformation, but with the temporal one. We change the definition of stationary states:
    [tex]\phi_n=\exp(i\epsilon_n t/\hbar )\bar{\phi}_n\;,[/tex][tex]\phi'_n=\exp(i\epsilon_n t\frac{1}{\hbar} +i \frac{e}{\hbar}\chi(t))\bar{\phi}'_n\;.[/tex]
    [itex]\bar{\phi}'[/itex] is time-independent as well as [itex]\bar{\phi}[/itex] Stationary Schroedinger equation yields[tex]\hat{H}'_0\bar{\phi}'=(\hat{H}_0-e\partial_t\chi)\bar{\phi}'=(\epsilon_n-e\chi(t)) \bar{\phi}' [/tex]We can see that all eigenenergies are shifted by a constant and phase of the wavefunction has changed. These changes do not change any physical quantities - we can always shift all energies by a constant.


    I do not see how to show general gauge invariance with respect to eigenvalue problem. I think the problem lies somewhere in the transition between time-dependent and stationary Schroedinger equations. Time can be separated easily only if our Hamiltonian is time-independent or if its time dependence is simple - there, it is clear how to define [itex]\bar{\phi}(r)[/itex] from [itex]\phi(r,t)[/itex]. For Hamiltonians that are time-independent in some gauge, there should be (IMHO) some transformation [itex]\phi(r,t)\to\bar{\phi}(r) [/itex] in any gauge. This transformation should change eigenvalues only by a constant. It is however difficult to find it.

    Why do I think there should be always such transformation? Because stationary states (=energy eigenstates if there is no deneneracy) are something physical. It's not only matter of formalism - if one prepares such state, probabilities of finding particle in space don't change, which is measurable property independent of the formalism. We must therefore be able to find these states even with complicated time-dependent [itex]H_0'(t)[/itex] which we get from general gauge transformations. It's only not clear how.
     
    Last edited by a moderator: Apr 26, 2017
  12. Nov 17, 2011 #11
    Just to summarize my previous post - it seems to me that the problem is probably not in Born's rule, but in the procedure of finding "stationary states" for time-dependent Hamiltonians. We have such procedure if the Hamiltonian is time independent (finding its eigensystem and adding the time-dependence, which is trivial). In such cases, the [itex]\hat{H}_0[/itex] can be chosen gauge-invariant and there is no problem if the gauge doesn't depend on time. If we choose gauge such that [itex]\hat{H}_0(t)[/itex] depends on time, we cannot use standard eigenproblem formulation (its not clear how to introduce relation between [itex]\phi(r,t)[/itex] and [itex]\bar{\phi}(r)[/itex]) to find the stationary states. I wonder if there are some articles about finding stationary states for time-dependent Hamiltonians.
     
  13. Nov 18, 2011 #12

    tom.stoer

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    I think you are right.

    Starting from a time-indep. Hamiltonian it's clear that we can solve the time-indep. Schrödinger equation for time-indep.eigenvalues.

    Applying a time-dep. gauge transformation means that we must use the time-dep. Schrödinger equation afterwards. But of course we can omit this b/c we already have constructed the eigenstates for the time-indep. equation and instead of solving the time-dep. equation we can simply apply the gauge transformation to the old eigenstates.

    If we know that the time-dep. problem has been generated by applying a time-dep. gauge trf. to a time-indep. problem everything is clear.

    But the other way round is difficult. Lookgn at an arbitrary time-dep. problem one could try to find a (time-dep.) gauge trf. to generate a time-indep. problem. If this is possible everything is fine again. If one can show that such a time-indep. gauge trf. does not exist one has to deal with the full complexity of the time-dep. problem.

    The problem with the proof is that one as to start with a time-dep. Hamiltonian w/o knowing its eigenstates i.e. before solving the Schrödinger equation and has to show that one single transformation (i.e. one single phase factor which is independent of the state) removes the time dependency from all states simultaneously!
     
  14. Nov 18, 2011 #13
    I think you can't simply take classical [itex]H_0[/itex] in arbitrary gauge and make it a quantum Hamiltonian [itex]\hat{H}_0[/itex]. The rule "take classical Hamiltonian and change classical quantities for QM operators" is not an exact procedure (because of the ordering problem). I think that [itex]H_0[/itex], [itex]A[/itex], [itex]\phi[/itex] has to be quantized in some gauge, the result must be compared with experiment and if it fits, then we found the correct quantum theory. Once you have it, it must be gauge-independent (which I think probably is even with respect to the eigenvalue problem, as I tried to show in the previous posts).

    Gauge invariance and correspondence principle:
    I'll ilustrate that if there is no ordering problem, one can probably choose arbitrary gauge on example of http://en.wikipedia.org/wiki/Particle_in_a_box" [Broken]. You showed me this example, I calculated it to end, so here is the solution:

    Hamiltonian of a free particle if (in general) [itex]H_0=\frac{(p-q A(r,t))^2}{2m}+\phi(r,t)[/itex]. We might ask what gauge we will choose for the solution of this problem. Let us exclude all time-dependent Hamiltonians (because there we do not know how to find stationary states). For free particle, this means [itex]\phi(r,t) = \phi_0[/itex]. [itex]A(r,t)[/itex] could be any field with zero rotation independent of time (which we excluded). However, in fact we cannot take [itex]A(r,t)[/itex] even r-dependent, because then there would be the ordering problem in our Hamiltonian. So let us show, that Hamiltonian
    [tex]\hat{H}_0=\frac{(\hat{p}-q A_0)^2}{2m}+\phi_0[/tex]
    leads to the same result for all [itex]A_0[/itex] and [itex]\phi_0[/itex]:

    The stationary Schroedinger's Eq. is:
    [tex]-\frac{\hbar^2}{2m}\psi''(x)+i\frac{q A_0\hbar}{2m}\psi'(x)+\left(\phi_0+\frac{q^2 A_0^2}{2m}\right)\psi(x)=e \psi(x)[/tex]with boundary condition given by the infinite potential at [itex]\psi(0)=0[/itex] and [itex]\psi(1)=0[/itex]. General solution with boundary condition [itex]\psi(0)=0[/itex] is
    [tex]\psi(x)=C \exp(i\frac{q A_0 x}{2\hbar})\sinh(\frac{x\sqrt{Z}}{2\hbar})[/tex]with[tex]Z=3q^2 A_0^2+8m(\phi_0-e).[/tex]Only way how to get non-zero solution satisfying [itex]\psi(1)=0[/itex] is for [itex]Z<0[/itex] using identity [itex]-i\sinh(ix)=\sin(x)[/itex]:
    [tex]\psi(x)=C' \exp(i\frac{q A_0 x}{2\hbar})\sin(\frac{x\sqrt{-Z}}{2\hbar}),[/tex]which leads to quantization
    [tex]e_n=\frac{3q^2A_0^2}{8m}+\phi_0+\frac{n^2 \pi^2 \hbar^2}{2m}.[/tex]Again, the differences of eigenenergies are independent of the chosen gauge. (But not their absolute values!)

    This probably suggests that you are free to choose gauge for quantization unless you run into the ordering problem or the problem of determining stationary states of time-dependent Hamiltonian.
     
    Last edited by a moderator: May 5, 2017
  15. Nov 21, 2011 #14
    Btw. when we do the gauge transformation
    [tex]A'=A+\nabla \chi,\; \varphi'=\varphi-\frac{1}{c}\frac{\partial\chi}{\partial t}, \; \Psi'=U\Psi=\exp\left(\frac{ie}{c\hbar}\right)\Psi\;,[/tex]operator related to every measurable quantity (for example Hamiltonian) http://electron6.phys.utk.edu/qm2/modules/m5-6/gauge.htm" [Broken]
    [tex]A=U^\dagger A' U\;.[/tex]Therefore the eigenproblem
    [tex]H\psi_n=\varepsilon_n'(t)\psi_n[/tex]transformes as
    [tex]H'\psi'_n=\varepsilon_n'(t)\psi'_n\implies[/tex][tex]UHU^\dagger U\psi_n=\varepsilon_n'(t) U\psi_n\implies[/tex][tex]H\psi_n=\varepsilon_n'(t)\psi_n[/tex]and [itex]\varepsilon_n'(t)=\varepsilon_n[/itex].

    This means that after the gauge transformation, the eigenvalues are unchanged even with time-dependent Hamiltonian, am I right? So the last question remains if I can perform quantization in every gauge, where is no operator-ordering problem, right?
     
    Last edited by a moderator: May 5, 2017
  16. Nov 21, 2011 #15

    tom.stoer

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    Of course; this is the central message of gauge-invariance.

    Afaik in QM with classical el.-mag. field there is no ordering ambiguity (better: a gauge-trf. does not introduce an additional ordering ambiguity for x and p).

    To make clear what I mean: for a simple one-dim. problem (e.g. harmonic oscillator) w/o gauge fields one could study V(x) ~ x2 and V'(x) ~ p x2 p-1; this ambiguity has nothing to do with gauge symmetry. Gauge symmetry itself is a well-defined unitary operator acting on the standard QM Hilbert space and does not induce such an ambiguity.
     
  17. Nov 21, 2011 #16
    OK, but this is still not so easy. For example if I choose [itex]\chi=f(t)[/itex], Hamiltonian [itex]H=\frac{p^2}{2m}+\varphi[/itex] tranformes into [itex]H'=\frac{p^2}{2m}+\varphi - f'(t)[/itex]. Since Hamiltonian commutes with time, [tex]H=U^\dagger H' U=H'\;[/tex]is not true and the eigenvalues get changed! I showed above that only by a constant, so energy differences do not change, but how to show this for gauge transformations that mix time and space? (It seems that statement that for operator of measurable quantity, [itex]A=U^\dagger A' U[/itex] must be satisfied is not true - and I thought it was solved :smile: . Or maybe it is because energy is not measurable quantity - energy difference is.)
    You are right, nice example - even with free particle we can get into ordering problems. One just has to accept that quantization is simply not an exact procedure and at the end only experiment can decide whether we have the correct Hamiltonian. (Jano L.'s question was, I think, why do I choose Hamiltonian [itex]H[/itex] and not [itex]H'[/itex], for quantization. Once I already have the correct Hamiltonian on Hilbert space, the problem is gauge-invariant, of course. But its not clear why should I do the quantization in one gauge or another. But again, it's not an exact procedure, so we are anyway just guessing.)
     
    Last edited: Nov 21, 2011
  18. Nov 21, 2011 #17

    strangerep

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    From post #14,
    In the last bit, did you mean [itex]\exp\left(\frac{ie\chi}{c\hbar}\right) \Psi[/itex] ?
     
  19. Nov 21, 2011 #18

    strangerep

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    Umm, how do you reach that conclusion? We'd also have [itex]\Psi = U^\dagger \Psi'[/itex].
     
  20. Nov 22, 2011 #19

    tom.stoer

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    Whenever you want to study arbitrary gauge transformation you have to use the full, time-dependent Schrödinger equation. Even for time-dep. gauge trf. the overall structure of this equation ensures that NO gauge trf. changes energy eigenvalues, probability densities or currents.

    Gauge symmetry is an intrinsic redundancy of the formalism and does never result in observable consequences.
     
  21. Nov 22, 2011 #20
    Yes, sorry.
    I think it should be obvious, that if Hamiltonian [itex]H=\frac{p^2}{2m}+\varphi(r)[/itex] has eigenvalues [itex]\varepsilon_n[/itex], Hamiltonian related by gauge transformation [itex]H'=\frac{p^2}{2m}+\varphi(r) - \frac{1}{c}\chi'(t)[/itex] will have eigenvalues [itex]\varepsilon_n'=\varepsilon_n - \frac{1}{c}\chi'(t)[/itex]. Sure, the eigenfunctions get transformed as well, but the eigenvalues get certainly changed. So the question is: if, for transormation [itex]\chi(t)[/itex], the eigenvalues get changed by a constant (which is all right), how do we know that in the general transformation [itex]\chi(r,t)[/itex], the eigenvalues' differences do not change as well?

    How do you show that time-dependent gauge transformation preserves the energy eigenvalues? (Or at least their differences.) Thanks.
     
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