Why is the phase of a matter wave not Galilean invariant? And what does this say about the Schrödinger equation?

Question

Matter waves are not Galilean Invariant

Consider a non-relativistic freely-propagating matter wave in an inertial frame $\Sigma'$ moving along the $x'$-direction with kinetic energy $E'=1/2m_0v'^2$, where $m_0$ is the rest mass of the particle and $v'$ the group velocity, wavelength $\lambda'=h/p'$ and momentum $p'=m_0v'$. At a time $t$ the phase, for constant velocity, is then given in the Schrödinger theory by $\Phi'(x',t)=\left(p'x'-E't\right)/\hbar$. Now consider the same situation, however, from the point of view of the inertial frame $\Sigma$ where $\Sigma'$ is moving at velocity $u$ relative to $\Sigma$, in Galilean relativity one transforms the coordinates as \begin{align} x'(t)&=x(t)-ut\label{eq:trans1}\tag{1}\\ v'&=v-u\label{eq:trans2}\tag{2}\\ t'&=t\tag{3} \end{align} The same phase expressed now from the $\Sigma$ coordinates is then \begin{align} \Phi'(x',t)=m_0\left(vx-ux-\tfrac{1}{2}v^2t+\tfrac{1}{2}u^2t\right)/\hbar\tag{4}. \end{align} However, we should just as equally be able to describe the same situation directly in the inertial frame $\Sigma$, which gives $\Phi(x,t)=m_0\left(vx-\tfrac{1}{2}v^2t\right)/\hbar$. This gives a relation between the phases \begin{equation}\tag{5}\label{eq:phase_trans_matter} \Phi'(x',t)=\Phi(x,t)-\frac{m_0}{\hbar}\left(ux-\tfrac{1}{2}u^2t\right). \end{equation} For classical waves, $\Phi'(x',t)=\Phi(x,t)$, this means Eq.(\ref{eq:phase_trans_matter}) shows us that under a Galilean boost, matter waves do not transform in a Galilean invariant way. In the Schrödinger theory the Fourier components of the wavefunction $\Psi(x,t)$ will all transform by Eq.(\ref{eq:phase_trans_matter}), this means that any solution $\Psi=e^{i\Phi}\psi$ to the Schrödinger equation, for free or trapped particles, a corresponding solution written with respect to the inertial frame $\Sigma$ can be written as \begin{equation}\label{eq:psi_G_inv} \Psi(x,t)=\Psi'(x',t)e^{i\frac{m_0}{\hbar}\left(ux-\tfrac{1}{2}u^2t\right)}\tag{6}. \end{equation} At first sight, it might be miss interpreted that this result is insignificant due to the fact that a system's total state vector being multiplied by a constant phase factor, doesn't change the outcome of a physical scenario. However, if Eq.(\ref{eq:psi_G_inv}) describes an individual component of the total state vector then it can have empirical significance.

The Sagnac Effect for Matter waves

The empirical significance can be seen as follows. The above discussion is also valid for the case of a one-dimensional path around a loop of radius $r_0$ and area $A=\pi r_0^2$. We can then consider two counter-propagating matter waves $\Psi'_\pm$ moving around a loop fixed in the now non-inertial frame $\Sigma'$ in time $T$ with equal and opposite angular velocities $\pm\omega'_p=\pm v'/r_0$, where $\omega'_pT=2\pi$. By substituting in $x'_\pm=\pm 2\pi r_0$, the experimenter in the rotating lab frame infers a zero phase difference from the two counter-propagating matter waves upon completing the full loop. However, from the inertial frame $\Sigma$ it is known that $u=\Omega r_0$, where $\Omega$ is the angular velocity of $\Sigma'$ and the correction factor Eq.(\ref{eq:psi_G_inv}) $\exp(im_0\left(\pm 2\pi \Omega r_0^2+\Omega^2t/2r_0^2\right)/\hbar)$ must be included for each matter wave. This then leads to an accumulated phase difference for a single trip around a loop of \begin{equation}\label{eq:seffect} \Phi = \frac{4m_0A\Omega}{\hbar}\tag{7}. \end{equation} Apart from a relativistic correction factor $1/\sqrt{1-\Omega^2 r_0^2/c^2}$, where $c$ is the speed of light, and which accounts for the relativistic mass increase, Eq.(\ref{eq:seffect}) is the same phase as predicted by relativity using the relativistic law of velocity composition for `classical' waves propagating around a loop and is called the Sagnac effect. Using the Galilean law of velocity composition for classical waves predicts zero phase difference.

Questions

1. Why aren't matter waves Galilean invariant?

2. What does it say about the Schrödinger equation and matter waves that effects attributable to special relativity can be predicted by a non-relativistic theory?

Answer 1

You discovered the substance of the so-called Bargamann superselection rule of the mass. The subsequent step should be to realize that, if the mass is an eigenvalue of some mass operator, then

[Bargmann's superselection rule] coherent superpositions of states with different values of the mass aren't permitted in non-relativistic QM.

(the same fact is true for the electric charge in general). Otherwise you would see an internal phase of the superposition when acting on the state with a symmetry operation corresponding to the trivial Galileian transformation (using the fact that the representation is unitary projective but the multipliers depend on the mass).

Here is a sketch of proof. Consider the Galileo group $G$. The quantum representations $G \ni g \mapsto U^{(m)}_g$ are projective unitary, namely $$U^{(m)}_gU^{(m)}_{g'} = \omega_m(g,g') U_{gg'}\:,\tag{1}$$ where the unit-modulus complex numbers $\omega_m$ have the following properties.

(a) They cannot be removed by changing an arbitrary phase (remember that states are unit vectors up to phases) in the definition of $U_g^{(m)}$ $$U_g \to V_g := \chi_g U_g\:.$$ In particular $G$ does not satisfy a sufficient co-homological condition due to Bargamnn that asseres that the numbers $\omega_m(g,g')$ can be removed by re-defining the $U_g$ with phases.

(b) $\omega_m(g,g')$ depends on the mass $m$.

Suppose that a state is a coherent superposition of states with different masses $m\neq m'$: $$\psi = a\psi_m+ b\psi_{m'}$$ The global action of the Galieleo group produces $$U_g\psi = aU^{(m)}_g\psi_m + bU^{(m')}_g \psi_{m'}$$ so that, $$U_{g^{-1}}U_g\psi = aU^{(m)}_{g^{-1}} U^{(m)}_g\psi_m + bU^{(m')}_{g^{-1}} U^{(m')}_g \psi_{m'} = a\omega_m(g,g^{-1}) \psi_m + b\omega_{m'}(g,g^{-1}) \psi_{m'}\:.$$ The vector on the left-most side should be the initial vector up to a phase since, as already pointed out, (pure) states are unit vectors up to phases. However, by direct inspection, one sees that $$\omega_m(g,g^{-1}) \neq \omega_{m'}(g,g^{-1})\quad \mbox{if $m \neq m'$}$$ so that the right-most side does not represent the initial state unless either $a=0$ or $b=0$. $\Box$

The ultimate physical reason for the appearance of non-trivial phases depending on $m$ in front to the transformed wavefunction (but also in (1)) is the same as in Hamiltonian mechanics. Galileo's group is defined in the space of positions and velocities for all bodies independently of their masses, as it is a kinematical notion regarding reference frames. However when you have to extend its action to the space of the states, where the momentum shows up, you need a further crucial piece of information to represent the action of the group. That is the mass of the system, since $p \to p + mV$ and $m$ is not part of Galileo transformations. In position representation, the argument of $\psi$ transforms according to the standard action if the Galileo group, no information about the mass is necessary. Therefore the only place where it can take place is in an external factor. A phase, just to preserve the total probability. However not a trivial constant phase which could be removed (states are unit vectors up to phases), but a phase depending on coordinates to keep the information about the mass. That is just an intuitive explanation of the mathematical results you found out. In summary, matterwaves transform through a non-trivial way under the action of Galileo's group. There is no reason why they should be invariant or why they should transform as scalars: they are not scalars! However they keep to satisfy the (free) Schroedinger equation as physically expected. If $\psi$ is the state at time zero, the active boost transformation at time $t$ is $$\psi'(t,x):=e^{ivK(t)}\psi(t,x)$$ but $$\psi(t,x) = (e^{-itH}\psi)(x)$$ and, just because the unitary representation of the active boost transformation ($x\to x+vt$) represented by the one-parameter group $e^{ivK(t)}$ of unitary operators is a (time-dependent dynamical) symmetry, $$e^{ivK(t)}e^{-itH} = e^{-itH} e^{ivK(0)}\:.$$ In summary $$\psi'(t,x) = e^{-itH} e^{ivK(0)} \psi(x)$$ so that it satisfies the Schroedinger equation with respect to the initial $H$ as I said.

The passage to the relativistic realm is quite difficult because (1) there are many different wave equations in relativistic quantum mechanics which admit the Schroedinger equation as non relativistic limit; (2) some of them are non-local when requiring the validity of a probabilistic interetation (i.e. they are pseudodifferential equations); (3) the notion of position observable (and position representation) turns out to be quite subtle... For these reasons, I do not think that relativistic effects can be explained or predicted in non-relativistic quantum mechanics also because the nature of the unitary representations of the Poincaré group is completely different: there are no annoying numbers $\omega(g,g')$ there. Or if there are, they can be removed by defining the operators $U_g$ with further phases as I wrote above.

Answer 2

First of all, since Galilean invariance is a physical symmetry (dynamical) of the system, which is not a gauge redundancy (non-dynamical), one should not expect that a quantum state should be invariant under a Galilean boost at all. The physical symmetry group acts on a physical state in the Hilbert space by transforming it into other physical states. However, physical states should form a trivial representation of the gauge group. The action of a gauge group on any physical state in the Hilbert space leaves it unchanged, since physics should be invariant under gauge transformations.

This is similar to the case of 1-particle states in relativistic quantum field theory, where the Poincare transformation $(\Lambda,a)$ acts on a state $|p,\sigma\rangle$ of momentum $p$ as a unitary operator $U(\Lambda,a)$. Here, $\Lambda$ is a Lorentz transformation and $a$ is a spacetime translation, and $\sigma$ stands for some other possible labels of the quantum state (such as spin). Since for any Poincare transformation $U(\Lambda,a)$, one can always write it as a product

$$U(\Lambda,a)=U(\Lambda,0)U(\mathbb{I},\Lambda^{-1}a),$$

it is sufficient to understand the following identities

$$U(\mathbb{I},a)|p,\sigma\rangle=e^{ip\cdot a}|p,\sigma\rangle,$$

$$U(\Lambda,0)|p,\sigma\rangle=\sum_{\sigma^{\prime}}C_{\sigma\sigma^{\prime}}(\Lambda,p)|\Lambda p,\sigma^{\prime}\rangle,$$

where the coefficients $C_{\sigma\sigma^{\prime}}(\Lambda,p)$ are to be determined by the mass of the particle. You should find more details from Wigner's classification of non-negative energy unitary irreducible representations of the Poincare group.

In quantum mechanics, there is a theorem about physical symmetry, known as the Wigner's theorem, which says that for every symmetry $g\in G$ acting on the projective Hilbert space $\mathbb{P}\mathcal{H}$, there exists either a unitary or an anti-unitary operator $U(g)$ acting accordingly on the Hilbert space $\mathcal{H}$. The consequence of this theorem is that two consecutive actions $U(g)$ and $U(h)$ on the Hilbert space can only be determined by $U(gh)$ up to a phase factor, i.e there exists a function $\omega:G\times G\rightarrow U(1)$ such that

$$U(gh)=\omega(g,h)U(g)U(h).$$

In other words, in QM one should be interested in the central extension of the symmetry group. In your case, since you are interested in the action of Galilei group, there is some special property of its Lie algebra that plays a key role in physics. Let me denote the Lie algebra of the Galilei group in three dimensional Euclidean space as $\mathfrak{gal}(3)$, with the following commutation relations

$$[J_{i},J_{k}]=\sum_{k}\epsilon_{ijk}J_{k},\quad [J_{i},K_{j}]=\sum_{k}\epsilon_{ijk}K_{k},\quad [J_{i},P_{j}]=\sum_{k}\epsilon_{ijk}P_{k},\quad [K_{i},H]=P_{i},$$

then one can compute its second cohomology group, which turns out to be non-trivial. To be specific, one finds

$$\mathrm{H}^{2}\left(\mathfrak{gal}(3),\mathbb{R}\right)\simeq\mathbb{R}.$$

One finds that the non-trivial central extension of the Galilei Lie algebra is parameterized by a real number, which corresponds to the mass of the particle in physics. We then have the central-extended Galilean Lie algebra

$$[J_{i},J_{k}]=\sum_{k}\epsilon_{ijk}J_{k},\quad [J_{i},K_{j}]=\sum_{k}\epsilon_{ijk}K_{k},\quad [J_{i},P_{j}]=\sum_{k}\epsilon_{ijk}P_{k},\quad [K_{i},H]=P_{i},\quad [K_{i},P_{j}]=m\delta_{ij},$$

where the central charge $m$ is the mass which parameterizes the central extension.

With the above commutation relations in mind, and notice that the operators $H$ and $P_{i}$ commutes, one can then define the eigenstates $|E,\vec{p}\rangle$ as the one-particle states in Galilean mechanics,

$$H|E,\vec{p}\rangle=E|E,\vec{p}\rangle,\quad P_{i}|E,\vec{p}\rangle=p_{i}|E,\vec{p}\rangle.$$

Then, using the commutator

$$[\vec{P},e^{i\vec{v}\cdot\vec{K}}]=m\vec{v}e^{-i\vec{v}\cdot\vec{K}},$$

one finds

$$\vec{P}\left(e^{i\vec{v}\cdot\vec{K}}|E,\vec{p}\rangle\right)=(m\vec{v}+\vec{p})\left(e^{i\vec{v}\cdot\vec{K}}|E,\vec{p}\rangle\right).$$

A similar computation shows that

$$H\left(e^{i\vec{v}\cdot\vec{K}}|E,\vec{p}\rangle\right)=\left(E+\vec{v}\cdot\vec{p}+\frac{1}{2}m\vec{v}^{2}\right)\left(e^{i\vec{v}\cdot\vec{K}}|E,\vec{p}\rangle\right).$$

This means that the Galilean boost shifts the energy and momentum of a particle at rest by

$$E^{\prime}=E+\frac{m\vec{v}^{2}}{2},\quad \vec{p}^{\prime}=m\vec{v},$$

which is exacly one should expect from classical mechanics.

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To Be Continued