4
$\begingroup$

The global $U(1)$ symmetry in Quantum Mechanics corresponds to the freedom to shift the phase of the wave function $$ \Psi \to e^{i\varphi} \Psi \, $$ and can be used to understand the conservation of electric charge.

Where, if at all, does this symmetry show up in alternative formulations of Quantum Mechanics, such as

  • the path integral formulation,
  • the phase space formulation,
  • the pilot wave formulation?
$\endgroup$
  • 1
    $\begingroup$ The global phase freedom in QM is not the same as the global $U(1)$ symmetry of electromagnetism! The former acts as $\psi\mapsto \mathrm{e}^{\mathrm{i}\varphi}\psi$ on all states, which the latter acts on states with electric charge $e$ as $\psi\mapsto\mathrm{e}^{\mathrm{i}e\varphi}\psi$. That is, the former is always a multiple of the identity, the latter one isn't if the space of states contains states of different charges. $\endgroup$ – ACuriousMind Oct 9 '18 at 16:43
4
$\begingroup$

This will only be a partial answer, i hope that is ok.

In Schrödinger quantum mechanics one usually works with complex vectors $\Psi$ which are normalized as $\|\Psi\| = 1$. This still leaves the freedom to choose a $U(1)$ phase. However, this is not a symmetry and in particular has nothing to do with conservation of electric charge.

1) If you recall the axioms of quantum mechanics, they say that physical states are represented by rays in Hilbert space, i.e. by complex one-dimensional subspaces $\mathbb{C} \Psi$. In calculations, it is often useful to choose a particular representative of this equivalence class $\Psi$ and formulate an evolution equation for this representative. However, the resulting paths of rays $\mathbb{C} \Psi(t)$, cannot depend on the choice of representative, since all vectors in one ray are physically indistinguishable. Hence the "symmetry" under phase rotations. If it would be absent, we would have made a mistake!

2) In many-particle quantum mechanics, a given Hamiltonian may conserve particle number or electric charge, or it might not. Denote by $N$ the operator has the eigenvalue $n$ when applied to a $n$-particle vector $\Phi$. Then the $U(1)$-symmetry associated to particle number conservation is

$ U(1) \ni e^{i \varphi} : \Psi \mapsto e^{-i N \varphi} \Psi $,

That is, it rotates all $n$-particle components of the wave-function with a different phase. This is not a change of representative in a ray and thus it is physically meaningful. Note how this looks in second quantization, i.e. introduce an operator $a(\psi)^\dagger$ with $\psi$ a single particle wavefunction and

$a(\psi)^\dagger \Omega = \psi$

where $\Omega$ is the Fock vacuum. Then the above $U_1$ symmetry transformation induces an action of operators by

$ U(1) \ni e^{i \varphi} : a(\psi)^\dagger \mapsto e^{i N \varphi} a(\psi)^\dagger e^{-i N \varphi} = e^{i \varphi} a(\psi)^\dagger$

3) In the path-integral formalism for $n$-body bosonic quantum mechanics, one often uses the representation in terms of creation and annihilation operators, that is, the action is a functional of the complex functions $\{a_i(t)\}_{i=1,2,\cdots n}$:

$S[a(t)] = \int_{t_i}^{t_f} \left[ \sum_i \overline{a}_i(t)\partial_t a_i(t) - H(\{a_i(t)\}_{i=1,2,\cdots n})\right] d t $

and the system will preserve the particle number if it is invariant under $U(1)$ phase rotations of the operators.

$\endgroup$
  • 1
    $\begingroup$ Small completion footnote: In the phase-space formulation one forsakes wavefunctions for bilinears of them, where such phases cancel automatically, ab initio, just as they cancel in density matrices. $\endgroup$ – Cosmas Zachos Oct 9 '18 at 14:24
1
$\begingroup$

In the pilot-wave formulation, the freedom to perform phase shits of the wave function $\Psi$ correspond to the freedom to add a total time derivative to the Lagrangian:

$$ L \to L + \frac{d \Lambda}{dt} . $$

This comes about, because to derive the equations of the pilot wave formulation, we use the ansatz $$ \Psi = R e^{iS} $$ in the Schrödinger equation.

This leads to two equations: the continuity equation and the Hamilton-Jacobi equation which tells us that we can interpret $S$ as the action.

Now, a phase shift of the wave function means $$ \Psi \to e^{i\varphi} \Psi $$ which means for $R$ and $S$: $$ R e^{iS} \to R e^{i\varphi} e^{iS} = R e^{iS + i\varphi } \, .$$ Therefore $$ R\to R$$ $$ S \to S +\varphi .$$

(In this sense, gauge transformations are a special type of the well-known canonical transformations, c.f. this paper.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.