1

I’ve read Ballentine where he derives the conserved observable operators (momentum, energy, ...) from symmetries of space-time.

Can I read up such a derivation in more detail somewhere else or even one for the Poincaré group?

And more importantly, the derivation uses gauge invariance of the wavefunction. Is there a derivation with the density matrix formalism where you wouldn’t need complex exponentials for gauge invariance?

Qmechanic
  • 201,751
Gere
  • 1,457

1 Answers1

2

Yes, there is:

A unitary 1-parameter group of transformation $U(s)$ satisfies $U(0)=1$, $U(-s)=U(s)^*$, and $U(s+s')=U(s)U(s')$ for all real $s,s'$. In the Heisenberg picture, $U(s)$ transforms an observable $A$ into the conjugate observable $A(s)=U(s)AU(-s)$, thereby preserving the spectrum of $A$.

The transformation is called a continuous symmetry of a quantum system if it preserves the Hamiltonian $H(t)$, i.e., $U(s)H(t)U(-s)=H(t)$ for all $s$, $t$. Differentiation with respect to $s$ gives $U'(s)H(t)U(-s)-U(s)H(t)U'(-s)=0$. Taking $s=0$ and introducing the generator $K=\dot U(0)/i\hbar$ of the symmetry, we find $K H(t)-H(t)K=0$. Thus $K$ commutes with the Hamiltonian at all times $t$. Conversely, if $K$ commutes with the Hamiltonian at all times, then $U(s)= e^{isK/\hbar}$ also commutes with the Hamiltonian at all times, so that $U(s)H(t)U(-s)=U(s)U(-s)H(t)=H(t)$. Thus a 1-parameter group is a group of symmetries iff its generator commutes with the Hamiltonian at all times.

A quantum system is called time invariant if the Hamiltonian $H$ is independent of time. In this case, $K=H$ trivially commutes with $H$, defining the symmetry group of time translations $U(t)=e^{itH/\hbar}$. Using this group, one can define for an arbitrary observable $A$ the timeshifted observable $A(t)=U(t)AU(-t)$, describing the time-dependence in the Heisenberg picture.

In the Schroedinger picture, one refers everything to time zero by writing time dependent expectations $\langle A\rangle_t=Tr\ \rho A(t)$ with respect to a fixed Heisenbeg state $\rho$ as expectations $\langle A\rangle_t=Tr\ \rho(t) A$. Using the definition of $A(t)$ and the properties of the trace, the resulting condition $Tr\ \rho A(t)=Tr\ \rho(t) A$ gives (for time independent systems) the formula $\rho(t)=U(-t)\rho U(t)$, with sign of $t$ opposite as in the Heisenberg picture! Differentiation gives $\dot\rho(t)=-U'(-t)\rho U(t)+U(-t)\rho U'(t)$, and since $U'(t)=iHU(t)/h_bar=iU(t)H/h_bar$, we find the quantum Liouville equation $\dot \rho(t) = -i/\hbar Tr\ [H,\rho]$. (In the time-dependent case, this equation takes the form
$\dot \rho(t) = -i/\hbar Tr\ [H(t),\rho]$.)

A quantum observable $A$ is called conserved if, for all states $\rho$, its expectation $\langle A\rangle_t$ is independent of time. Differentiation gives the condition $0=Tr\ \dot \rho(t) A = -i/\hbar Tr\ [H(t),\rho] A$, which is equivalent with $0=Tr\ [H(t),\rho] A = Tr\ H\rho A-Tr\ \rho H A = Tr\ \rho (HA-AH) =Tr\ \rho [H,A]$. Since this must hold for all $\rho$, the observable $A$ is conserved iff it commutes with $H$.

This proves Noether's theorem that the generator of a 1-parameter group of symmetries is conserved.

If you specialize this to the case where $A$ is the energy operator (the generator of time translations), a component of the momentum operator (the generators of space translations), a component of the angular momentum operator (the generators of rotations), you get from invariance under the Poincare group the conservation of energy, momentum, angular momentum.

Arnold Neumaier
  • 44,880
  • This seems to already assume QM framework? Ballentine derives representations in terms of position and momentum for the Galilei group. Can I do the same with the density matrix? Because at the beginning I don't even know that such of thing like $\hbar$ exists. It has to be derived. – Gere Mar 08 '12 at 09:45
  • I don't have Ballentine's book, hence cannot resolve this query. However, without assuming a QM framework you cannot derive anything, and I don't believe that Ballentine gets his result without assuming what amounts to a complete QM framework. See my answer in http://physics.stackexchange.com/a/22019/7924 for the little I need to assume; it is very unlikely that B. assumes less - though very likely that he builds things up from different asumptions, ending with the same formalism. – Arnold Neumaier Mar 08 '12 at 10:10
  • I'm new to this topics, but to me it seems B. is more fundamental. He starts with saying there is a density matrix and observables as operators. Averages are the trace. Then he works with pure states knowing that states are phase invariant. For the Galilei group he deduces all commutators and operators by representation theory. In general due to gauge there is one free variable [x,px]=[y,py]=[z,pz]=A which turns out to be $\hbar$ in experiment. That's the derivation. – Gere Mar 08 '12 at 14:20
  • He basically derives the form of the observables without prior knowledge about anything but the maths of the Galilei group (existence of density matrix and trace averages are postulated). He expresses all operators in terms of position and momentum operators while phase invariance allows for one free constant. – Gere Mar 08 '12 at 15:23
  • So what is the difference in quality to what I did? I didn't assume anything but the math of linear operators, which is fixed once you have a Hilbert space. Noether's theorem then follows from the definitions I gave. Once you plug in the concrete symmetry group (Galilei of Poincare), you get the corresponding conservation laws. Gauge considerations don't even enter. That $\hbar$ appears in my formulas is only to get the conventional units of the generators, and not essential for the argument. – Arnold Neumaier Mar 08 '12 at 15:32
  • Why does $U$ preserve the time derivative in the way shown? If that has to do with Hamiltonian formalism, then that's a postulate Ballentine doesn't need. Moreover "pluging in the concrete symmetry" is the key part and from the derivation I'm interested in. I hope to find it more in detail somewhere. It even seems that phase invariance of the way function actually plays a key role for the story of QM. – Gere Mar 08 '12 at 16:53
  • There is no differentiation with respect to $t$ in my argument involving $U$. - To plug things in you don't need any extra computation - I just meant specialize Noether's theorem (which I had proved) to the case of Galilei or Poincare symmetries. Thus this is trivial. - in the density matrix formalism, you don't need any phase invariance as a density matrix has no phase. Even for a pure state $\psi$, the phase automatically cancels when you form the corresponding density matrix $\rho=\psi\psi^*$. It is far mor natural to work with density matrices than with pure states. – Arnold Neumaier Mar 08 '12 at 17:04
  • In the first part you transform time differentiation to H which hasn't been proven yet. I want to deduce from basics and not use ready-made Hamiltonian framework. Hmm, I don't think you get my point here, but thanks. Your answer provides part of the explanation. – Gere Mar 08 '12 at 20:52
  • To start from the schroedinger equation and deduce the dynamics of the density matrix, or to start with the synamics of the density matrix and deduce the Schroedinger equation is completely equivlaent. You need to assume one form of dynamics and get from it the other. At theend you have both. - But the density version is the direct quantum analogue of the classical Liouville equation, whereas the Schroedinger equation has no motivation at all except that it works. – Arnold Neumaier Mar 09 '12 at 08:05
  • That's my point. It seems Ballentine deduces all of the above, including the form of H and it's equation, from the Galilei group. No dynamics or Liouville is assumed. Effectively, even Schrödinger is also deduced! The only assumption is about complex Hilbert space, phase invariance and Galilei group. That's why it's interesting. – Gere Mar 09 '12 at 09:41
  • let us continue this discussion in chat – Arnold Neumaier Mar 09 '12 at 10:49