# Difference between symmetry transformation and basis transformation

What is the difference between a basis transformation and a symmetry transformation in the Hilbert space of a quantum system?

By a basis transformation, I mean transforming from one orthonormal basis $\{|\phi_n\rangle\}$ to another $\{|\chi_n\rangle\}$. A state $|\psi\rangle$ in the Hilbert space can be expanded in these two bases as $$|\psi\rangle=\sum\limits_{n}C_m|\phi_m\rangle=\sum\limits_{i}D_i|\chi_i\rangle$$ where $\langle\phi_m|\phi_n\rangle=\delta_{mn}$ and $\langle\chi_i|\chi_j\rangle=\delta_{ij}$. The change of basis is a unitary transformation i.e., $$|\chi_n\rangle=U|\phi_n\rangle.$$

By a symmetry transformation, I understand a rotation (for example). How is that different from a basis transformation?

• What are the definitions of these two transformations you are working with, and what specifically about their difference is unclear to you? – ACuriousMind Apr 25 '18 at 17:07
• @ACuriousMind Edited – SRS Apr 25 '18 at 17:20
• You wrote down how a state can be expanded in two bases, but you didn't write down what you consider the actual transformation. – ACuriousMind Apr 25 '18 at 17:27
• Yes. But isn't rotation just one such basis transformation? I think I'm confusing. – SRS Apr 25 '18 at 17:29

## 1 Answer

Some comments probably related to your confusion:

1. Just writing a state in two different bases is not a transformation, you aren't doing anything to the state. A transformation is a non-trivial map from the Hilbert space to itself.

2. Given two different bases $\{\lvert \psi_i\rangle\}$ and $\{\lvert \phi_i\rangle\}$, the map $$U: H\to H, \lvert \psi_i \rangle\mapsto \lvert \phi_i\rangle$$ is a unitary operator with matrix components $U_{ij} = \langle \psi_i \vert \phi_j\rangle$ in the $\psi$-basis (compute this explicitly if you do not see it).

3. There are two different notions of symmetry in this context (see also this answer of mine:

The weaker one is that a symmetry is a transformation on states that leaves all quantum mechanical amplitudes invariant, this is a symmetry in the sense of Wigner's theorem which tells us that such transformations are represented by unitary operators.

The stronger one is that a symmetry is a symmetry in Wigner's sense that additionally commutes with time evolution, i.e. whose unitary operator commutes with the Hamiltonian.

• But a state does change under a basis transformation. Right? In your example, if the bases are related by $|\phi_i\rangle=U|\psi_i\rangle$, an arbitrary state $|\alpha\rangle$ changes as $|\alpha\rangle\to U^\dagger |\alpha\rangle$. This is what I find in Sakurai.@ACuriousMind – SRS Apr 25 '18 at 17:45
• @SRS If by "basis transformation", you mean the unitary transformation in my second point, then yes. If you simply mean writing a state in two different bases as you did in your question, then no. – ACuriousMind Apr 25 '18 at 17:49
• So for Weinberg, a symmetry transformation is just a basis transformation? Basis transformation also keeps quantum mechanical amplitudes unchanged. @ACuriousMind – SRS Apr 25 '18 at 19:58
• @SRS That's probably not true even if by "basis transformation" you mean the $U$ from my second point, which you still haven't clarified. I'm sure Weinberg also uses the dynamical notion of symmetry. "Weinberg's symmetry" in my linked answer just denotes the notion of symmetry in the passage the question there quotes. You always need to look at the context to see what exactly a symmetry is in the given context. – ACuriousMind Apr 25 '18 at 20:01
• By basis transformation, I mean a linear transformation between two orthonormal bases which are related by a unitary operator to preserve orthonormality of basis vectors. I explain here physics.stackexchange.com/questions/402066/… what I mean by a basis transformation. Note that both the symmetry transformation and the basis transformation are unitary, and QM amplitude preserving. – SRS Apr 25 '18 at 20:05