My professor told us that in quantum mechanics a transformation is a symmetry transformation if $$ UH(\psi) = HU(\psi) $$
Can you give me an easy explanation for this definition?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
7
In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion.
In this case, the equation of motion is the Schrödinger equation $$ i\hbar\frac{d}{dt}\psi=H\psi. \tag{1} $$ We can multiply both sides of equation (1) by $U$ to get $$ Ui\hbar\frac{d}{dt}\psi=UH\psi. \tag{2} $$ If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as $$ i\hbar\frac{d}{dt}U\psi=HU\psi. \tag{3} $$ which says that if $\psi$ solves equation (1), then so does $U\psi$, so $U$ is a symmetry.
For a more general definition of symmetry in QM, see
answered Apr 8, 2019 at 13:12
-
3$\begingroup$ This is a good answer but it brings to another question, why do we call symmetry this condition? $\endgroup$ Commented Apr 8, 2019 at 13:39
-
$\begingroup$ @SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve. $\endgroup$ Commented Apr 8, 2019 at 16:58
-
1$\begingroup$ @SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry? $\endgroup$ Commented Apr 8, 2019 at 21:48
-
$\begingroup$ @Vectornaut What if they answered yes to any of those? What would you say? $\endgroup$– KrupipCommented Apr 9, 2019 at 17:01
-
1$\begingroup$ @SimoBartz The word "symmetry" is used with related-but-different meanings. So are the words "system is invariant." For example, we could define a symmetry (or invariance) to be any transformation that doesn't change the model's physical predictions. By that definition, every unitary transformation -- if applied both to the state-vectors and to all of the observables -- would trivially be a symmetry. The answer I gave here only offered a relatively easy motivation for the specific definition that you showed in the OP. $\endgroup$ Commented Apr 11, 2019 at 13:02
$\begingroup$
$\endgroup$
What you have written there is nothing but the commutator. Consider for example the time evolution operator \begin{align*} U\left(t-t_{0}\right)=e^{-i\left(t-t_{0}\right) H} \end{align*} If $\psi\left(\xi_{1}, \dots, \xi_{N} ; t_{0}\right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies $\left[U\left(t-t_{0}\right), P\right]=0$ then also $$\left(P U\left(t-t_{0}\right) \psi\right)\left(\xi_{1}, \ldots, \xi_{N} ; t_{0}\right)=\left(U\left(t-t_{0}\right) P \psi\right)\left(\xi_{1}, \ldots, \xi_{N} ; t_{0}\right)$$ This means that the permuted time evolved wave function is the same as the time evolved permuted wave function.
Another example would be if you consider identical particles. An arbitrary observable $A$ should be the same under the permutation operator $P$ if one has identical particles. This is to say: \begin{align*} [A, P]=0 \end{align*} for all $P\in S_N$ (in permutation group of $N$ particles).
