1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Galilei Transformation Free Schrödinger Equation

  1. Oct 10, 2014 #1

    VVS

    User Avatar

    1. The problem statement, all variables and given/known data
    I am supposed to show that the free Schrödinger Equation is NOT kovariant under Galilei Transformation.

    2. Relevant equations
    We learned in Lectures that the Galilei Transformation can be written as:
    [itex]\vec{x'}=\hat{R}\vec{x}-\vec{a}-\vec{v}t[/itex] (1) or equivalently:
    [itex]\vec{x}=\hat{R^{-1}}\left(\vec{x'}+\vec{a}+\vec{v}t\right)[/itex] (2)

    3. The attempt at a solution

    I have no problem with evaluating the LHS of the time dependent Schrödinger equation.
    However I am not sure if the second part of this is needed or not correct.
    [itex]\frac{\partial}{\partial x_i}=\sum_k \frac{\partial x'_k}{\partial x_i}\frac{\partial}{\partial x'_k}+\frac{\partial t}{\partial x_i}\frac{\partial}{\partial t}[/itex]
     
  2. jcsd
  3. Oct 12, 2014 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    This is a very strange question, because non-relativistic quantum theory is of course Galilei covariant. I don't understand, why you want to derive [itex]t[/itex] with respect to [itex]x_i[/itex]. The Galilei trafo for time is [itex]t'=t-b[/itex]. Concerning time there are only time translations in non-relativistic physics (Newton's idea of an absolute time).

    You'll see that the wave function transforms non-trivially under Galilei boosts, which is an (Abelian) subgroup of the full Galilei transformations you wrote down. It's defined by [itex]\hat{R}=\mathbb{1}_3[/itex], [itex]\vec{a}=0[/itex], i.e., it's the change from one inertial frame to another one moving with velocity [itex]\vec{v}[/itex] with respect to the former.

    But then remember, what's observable in quantum theory. Then you'll see that observable quantities do not change under Galilei boosts (and also not under the full Galilei transformation).
     
  4. Oct 12, 2014 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Last edited by a moderator: May 7, 2017
  5. Oct 19, 2014 #4

    VVS

    User Avatar

    Hey, thanks for your attempts to help me. I solved the problem. The Schrödinger equation is not covariant under Galileitransformations, but can be made so by multiplying the wavefunction with a plane wave function.
     
  6. Oct 20, 2014 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Again: Non-relativistic quantum mechanics IS COVARIANT under Galileo transformations. Otherwise it wouldn't make sense!

    Take one free particle as the most simple case. Then the symmetry is realized as
    [tex]t'=t, \quad \vec{x}'=\vec{x}-\vec{v} t, \quad \vec{p}'=\vec{p}-m \vec{v}, \psi'(t',\vec{x}')=\exp(\mathrm{i} \Phi) \psi(t,\vec{x})=\exp(\mathrm{i} \Phi) \psi(t',\vec{x}'+\vec{v} t').[/tex]
    Now we have to check that we can find a real phase [itex]\Phi(t',\vec{x}')[/itex] such that [itex]\psi'[/itex] fulfills the Schrödinger equation in the new variables, if [itex]\psi(t,\vec{x})[/itex] fulfills it in the old variables. We have
    [tex]\mathrm{i} \hbar \partial_{t'} \psi'=\mathrm{i} \hbar \exp(\mathrm{i} \Phi) \left [\mathrm{i} \partial_{t'} \Phi(t',\vec{x}') + \vec{v} \cdot \vec{\nabla} \psi(t,\vec{x})|_{t=t',\vec{x}=\vec{x}'+\vec{v} t'} \right ].[/tex]
    Further we find
    [tex]\Delta' \psi'(t',\vec{x}') = \left [\mathrm{i} \Delta' \Phi(t',x') \psi(t',\vec{x}'+\vec{v} t') +2 \mathrm{i} \vec{\nabla} ' \Phi(t',\vec{x}') \cdot \vec{\nabla}' \psi(t',\vec{x}'+\vec{v} t') + \Delta ' \psi(t',\vec{x}'+\vec{v} t') + (\vec{\nabla}' \Phi)^2 \psi(t',\vec{x}'+\vec{v} t') \right ]\exp(\mathrm{i} \Phi)[/tex]
    Now you demand that
    [tex]\mathrm{i} \hbar \partial_{t'} \psi(t',\vec{x}') = -\frac{\hbar^2}{2m} \Delta' \psi(t',\vec{x}'),[/tex]
    provided that
    [tex]\mathrm{i} \hbar \partial_{t} \psi(t,\vec{x}) = -\frac{\hbar^2}{2m} \Delta \psi(t,\vec{x}).[/tex]
    Comparing the coefficients in front of [itex]\vec{\nabla} \psi[/itex] and [itex]\psi[/itex], you get the system of PDEs for the phase factors:
    [tex]\vec{\nabla}' \Phi(t',\vec{x}') = -\frac{m}{\hbar} \vec{v}, \quad \partial_{t'} \Phi(t',\vec{x}') = \frac{\hbar}{2m} \mathrm{i} \Delta' \Phi(t',\vec{x}') -\frac{\hbar}{2m} [\vec{\nabla}' \phi(t',\vec{x}')]^2.[/tex]
    From the first equation you get
    [tex]\Phi(t',\vec{x}')=-\frac{m}{\hbar} \vec{v} \cdot \vec{x} + \Phi_1(t')[/tex]
    and from the 2nd
    [tex]\dot{\Phi}_1(t')=-\frac{\hbar}{2m} \vec{v}^2 \; \Rightarrow \; \Phi=-\frac{m}{\hbar} \vec{v} \cdot \vec{x}' - \frac{\hbar}{2m} \vec{v}^2 t' + \phi_0.[/tex]
    Thus the transformation law for the Schrödinger wave function under Galileo boosts reads
    [tex]\psi'(t',x')=\exp\left [\frac{m}{\hbar} \vec{v} \cdot \vec{x}' - \frac{\hbar}{2m} \vec{v}^2 t' + \phi_0 \right ] \psi(t',\vec{x}'+\vec{v} t').[/tex]
    This means that up to a phase factor the wave function is a scalar under Galileo boosts and thus quantum theory is covariant under Galileo boosts, because the measurable quantities are all invariant under the Galileo transformation.

    The Galileo group is realized as a unitary ray representation. There is no sensible realization of the Galileo group as a unitary transformation. The mass is a non-trivial central charge of the Galileo algebra, and setting it to 0 does not lead to useful quantum dynamics. Massless particles make no sense in non-relativistic physics at all. For a deeper understanding, see

    Inönü, E., Wigner, E. P.: Representations of the Galilei group, Il Nuovo Cimento 9(8), 705–718, 1952
    http://dx.doi.org/10.1007/BF02782239

    Ballentine, Leslie E.: Quantum Mechanics, World Scientific, 1998
     
  7. Oct 22, 2014 #6

    VVS

    User Avatar

    Thanks for the derivation.
    Then there must be some confusion about the exact definition of covariance. Because our professor asked us to prove that the SCHRÖDINGER EQUATION is not covariant under Galilei Transformations. And you can actually show that if you take the Schrödinger equation in the moving system and perform the Galilei-Transformation then you yield the Schrödinger equation in the resting system, but with an additive term.
    Then he asked us in a follow up question to determine the factor by which we have to multiply the wave function in the resting system to make it Galilei covariant.
     
  8. Oct 23, 2014 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Well, it may be a question of definition, but what does "covariance" or "invariance" mean? It means that the physically relevant quantities in a theory are independent under a symmetry transformation. In quantum theory the wave function is not directly observable but its modulus squared is the probability distribution for the position of the particle and thus a measurable quantity. This probability distribution should be Lorentz covariant, because otherwise quantum theory would not be consistent with the space-time structure of Newtonian mechanics and wouldn't make sense within a "Newtonian" quantum theory.

    Together with the general structure of quantum theory, this implies that the wave function (or more generally the Hilbert-space vectors representing (pure) states) are only defined up to a phase factor, and a symmetry transformation is only determined up to phase factors. Thus, the true representatives of the pure states are not the (normalized) Hilbert-space vectors but the rays in Hilbert space, i.e., each normalized vector can be multiplied by phase factors without changing its physical meaning.

    Consequently a symmetry transformation is represented by either a unitary or antiunitary ray representation (Wigner-Bargmann theorem). This is of utmost importance. First, as you have seen in your exercise, the very valid Schrödinger theory (non-relativistic quantum theory) cannot be reformulated in a way that you have a true unitary representation of the Galilei group. Because the Galilei group is a Lie group, i.e., a continuous group, the transformation must be a unitary ray representation. Now you can ask, whether it's possible to find a truely unitary representation of the Galilei group on Hilbert space. Of course one can, but it turns out that it doesn't lead to a physically sensible dynamics.

    The second reason for the importance that rays and not vectors represent states is the well-established existence of half-integer spins. For half-integer spinor representations of the rotation group (which is a subgroup of the Galilei group) the rotation around [itex]360^{\circ}[/itex] leads to a phase factor [itex]-1[/itex], but that doesn't matter, because multiplying all vectors by the same phase factor is irrelevant. This of course further implies that there must not be superpositions of states from a half-integer spin and an integer-spin particle, because a rotation around [itex]360^{\circ}[/itex] would lead to a different state, because the former pieces multiyply by -1 but the latter don't change. This shouldn't happen, because a rotation around [itex]360^{\circ}[/itex] means that you've done nothing at all, i.e., it's the identity transformation within the physical Galilei group. Indeed, so far nobody has ever found a superposition of half-integer and ingeger-spin states.
     
  9. Oct 25, 2014 #8

    VVS

    User Avatar

    Thank you for your comment. I learnt not only from doing the exercise but also from your comments.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Galilei Transformation Free Schrödinger Equation
  1. Schrödinger equation (Replies: 2)

  2. Schrödinger equation (Replies: 3)

Loading...