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In classical mechanics:

If a Lagrangian L is preserved by an infinitesimal change in the state space variables qi -> qi + εKi(q) leads to only second order change in the Lagrangian: $$ 0 = \frac{dL}{d\epsilon} = \sum_i \left( \frac{\partial L}{\partial q_i}K_i + \frac{\partial L}{\partial \dot{q}_i} \dot{K}_i \right) = \frac{d}{dt}\left(\sum_i \frac{\partial L}{\partial \dot{q}_i} K_i \right) $$

Then we get our conserved momentum because the rate of change on the right side is 0.

In quantum mechanics, an observable A commuting with the Hamiltonian [H,A] = 0, corresponds to a symmetry of the time-independent Schrodinger equation Hψ = Eψ. How to we compute the conserved quantity related to A? In particular, what is the the conserved quantity associated with the identity operator?

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Hmm. Isn' the conserved quantity $A$ itself? –  Mariano Suárez-Alvarez Jan 22 '10 at 3:57
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It might be a little clearer to the asker to say that the expectation value of A in any state is conserved. Thus, the conserved quantity associated to the identity operator is 1. –  Aaron Bergman Jan 22 '10 at 4:03
    
I added a tag for symplectic geometry and a tag for quantum mechanics. (Surprisingly a tag for quantum mechanics didn't exist!) –  Kevin H. Lin Jan 22 '10 at 20:42
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2 Answers

In hindsight, Noether's theorem is a dramatic hint of quantum mechanics. Mariano is completely correct in his comment that the conserved quantity is $A$ itself, but it deserves a bit of explanation.

A classical probabilistic system is characterized by an algebra of random variables. You could consider the Boolean random variables, in which case the algebra is a $\sigma$-algebra $\Omega$. Or you could consider real or complex random variables; if you take the bounded ones then the algebra is $L^\infty(\Omega)$. In quantum probability, you have the same sort of thing, except that the algebra of bounded complex random variables is a non-commutative von Neumann algebra. One choice with special properties is the algebra $\mathcal{B}(\mathcal{H})$ of all bound operators on a Hilbert space $\mathcal{H}$.

The special property of $\mathcal{B}(\mathcal{H})$ is that all automorphisms are inner, so that any symmetry $A$ of a quantum dynamical system is necessarily also a random variable that you can measure. This does not happen classically, nor even for other non-commutative von Neumann algebras. Even without writing down a time-independent Schrodinger equation, it makes Noether's theorem trivial, because the symmetry $A$ must be conserved if you interpret it as a quantum random variable. Unlike in the classical case, $A$ doesn't even need to generate or come from a continuous group action.

For example, the parity operator (which negates all three coordinates of space) is a conserved quantity of electromagnetism, so it leads to a (two-valued) conserved quantity in quantum electrodynamics which is also called parity. The discrete symmetry also exists classically as a symmetry of Maxwell's equations (if you are careful to negate the magnetic field vectors twice), but the classical Noether's theorem doesn't apply.

Anyway, the identity operator is the trivial random variable that is always 1, as Aaron says.

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Is there a quantum version of the algebra of Boolean random variables? Or should that be its own question? –  Theo Johnson-Freyd Jan 22 '10 at 6:18
    
A quantum Boolean is a self-adjoint projection (or if you like, idempotent) in the von Neumann algebra $\mathcal{M}$. The most reasonable way to put them into an algebra is just to say that they generate $\mathcal{M}$. For instance, unlike in the classical case, they are not closed under multiplication. They do also have join and meet operations, leading to the theory of quantum logic. However, quantum logic hasn't been nearly as interesting or useful as quantum probability. I do not know quantum logic axioms that yield von Neumann algebras. –  Greg Kuperberg Jan 22 '10 at 6:32
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Is "the algebra B(H) of all random variables on a Hilbert space H" supposed to read "the algebra B(H) of all bounded operators on a Hilbert space H"? If not, what is a random variable on a Hilbert space? –  Vectornaut Jun 3 '11 at 6:54
    
@Vectornaut - You are right, and I fixed it. –  Greg Kuperberg Jun 3 '11 at 10:06
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As already mentioned, the conserved quantity is A itself.

I'll elaborate a little to see the full analogy between the classical and quantum mechanical case.

Let's assume that the lagrangian depends on the coordinates $q_i\in R^n$ and their first derivatives. The classical hamiltonian $H(p,q)$ which is a function on the phase space $T^{*}R^n$ can be obtained through the Legendre transform.

Now if a transformation preserves the Lagrangian, you will see that there exists a function $A(p,q)$ on $T^{*}R^n$ whose canonical Poisson bracket with the Hamiltonian vanishes: $\{A(p,q), H(p,q)\} = 0$. When you implement this transformation on the coordinates, you get the transformation law you started with: $\{A(p,q),q_i\} = K_i(q)$,

Thus, what Noether theorem really does is to allow you compute a function $A(p,q)$ which canonically generates the transformation you started with. This function is conserved under the classical evolution since its Poisson brackets with the hamiltonian vanish.

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I fixed the braces (curly parentheses). Because of an interaction between Markdown and JSMath, you need to use two backslashes before a brace to get it to display. –  Theo Johnson-Freyd Jan 22 '10 at 6:17
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