Noether's theorem in quantum mechanics

Question

In classical mechanics:

If a Lagrangian $\mathcal{L}$ is preserved by an infinitesimal change in the state space variables $q_i \to q_i + \varepsilon K_i(q)$, this leads to only second order change in the Lagrangian: $$ 0 = \frac{d\mathcal{L}}{d\varepsilon} = \sum_i \left( \frac{\partial \mathcal{L}}{\partial q_i}K_i + \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \dot{K}_i \right) = \frac{d}{dt}\left(\sum_i \frac{\partial \mathcal{L}}{\partial \dot{q}_i} K_i \right). $$

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, i.e. with $[\hat{H},A] = 0$, corresponds to a symmetry of the time-independent Schrödinger equation $\hat{H}\Psi = E \Psi$. How do we compute the conserved quantity related to $A$? In particular, what is the conserved quantity associated with the identity operator?

Answer 1

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.

Answer 2

(The laws of physics are supposedly invariant under time translations, so I hope the lateness of this answer can be forgiven.)

Both the question and the previous answers seem to have overlooked the fact that, in quantum mechanics, symmetries correspond to unitary operators on Hilbert space while observables (such as conserved charges) correspond to self-adjoint operators.

Supposing the Hilbert space to be finite-dimensional for simplicity, given any unitary matrix $A$, there exists an hermitian matrix $Q$ such that $A = e^{iQ}$, but $[H,A]=0$ (i.e. that $A$ is a symmetry of the hamiltonian) does not necessarily imply that $[H,Q]=0$ (i.e. that $Q$ is a conserved charge). Counterexamples may readily be found by considering $Q = \begin{pmatrix} 0 & -i\pi \\ i\pi & 0 \end{pmatrix}$, for which $A = -\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ commutes with any $H$.

But suppose, as for Noether's theorem in classical mechanics, that we have a smooth family $\theta \mapsto A_\theta$ of symmetry transformations such that $A_0$ is the identity transformation. Then the exponential map furnishes us with a $Q$ such that $A_\theta = e^{i\theta Q}$ and the Baker-Campbell-Hausdorff lemma in the form $[H,A_\theta] = i\theta [H,Q] + O(\theta^2)$ guarantees that $Q$ is conserved.

Answer 3

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.

Answer 4

I) In Heisenberg formulation of quantum mechanics you directly have that the observable $A$ is constant since $$\frac{d}{dt}A(t) = \frac{i}{\bar{h}}[\hat{H},A] $$

II) In the schrodinger formulation with the wave function, $A$ and $\hat{H}$ commute and then can be diagonalised simultanuously : There exists an orthogonal familly of eigenvector $(\phi_i)_{i\in \mathbb{N}}$ such that $\hat{H}\phi_i = \epsilon_i \phi_i$ and $A\phi_i = a_i\phi_i$ where $\epsilon_i$ and $a_i $ are the eigenvalues of $\hat{H}$ and $A$. Decomposing the wave function in this basis and with the Schrodinger equation $$ \phi(t)=\sum_i c_i(t)\phi_i \qquad i\frac{d}{dt}\phi(t) = \hat{H}\phi(t)$$ we have $c_i(t)=e^{-i\epsilon t}c_i(0)$ for all $t$. In particular $|c_i(t)|^2=|c_i(0)|^2$ which meens that the norm of the wave function restricted to any eigenspace of $A$ is constant. This also implies that for any real function $f$, $\langle \phi(t),f(A) \phi(t)\rangle$ is constant. Finally when performing the measure of the system the probability $\mathbb{P}(A=a_i)$ is also constant.