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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?

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1 Answer 1

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.

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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. –  Gerenuk Mar 8 '12 at 9:45
    
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 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 8 '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. –  Gerenuk Mar 8 '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. –  Gerenuk Mar 8 '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 8 '12 at 15:32
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