# Invariance of commutator relations under change of basis

Consider the following:

Let for operators $$\hat A$$ and $$\hat B$$ the following commutation relation holds:

$$[\hat A,\hat B]=\hat C \tag{1}$$

and now we know that this relation holds,

$$[\hat A',\hat B']=\hat C' \tag{2}$$

where,

$$\hat A'=T_{_{X\leftarrow Y}}A$$

$$\hat B'=T_{_{X\leftarrow Y}}B$$

$$\hat C'=T_{_{X\leftarrow Y}}C$$

(Since commutation relations do not depend on a change of basis )

Now, it's easy to prove this for a Unitary transformation $$U$$ (which is nothing but a change of basis) as follows:

$$[\hat A',\hat B']=\hat A'\hat B'-\hat B'\hat A'=UAU^{\dagger}UBU^{\dagger}-UBU^{\dagger}UAU^{\dagger}=U(AB-BA)U^{\dagger}=UCU^{\dagger}=C'$$

where now, (I think this is where the problem is$$^1$$ )

$$\hat A'=UAU^{\dagger}$$

$$\hat B'=UBU^{\dagger}$$

$$\hat C'=UCU^{\dagger}$$

Here U (which is nothing but a transformation/transition matrix) is getting applied on both ends since A is an operator here on a ket of Hilbert space as opposed to the transition matrix which just gets pre-multiplied.$$^{2}$$

When I try to prove $$(2)$$ from $$(1)$$ for my previous case involving transition matrix this happens;

$$[\hat A',\hat B']=\hat A'\hat B'-\hat B'\hat A'=T_{_{X\leftarrow Y}}AT_{_{X\leftarrow Y}}B-T_{_{X\leftarrow Y}}B T_{_{X\leftarrow Y}}A$$

I am stuck here and I think the resolution is related to $$^{1}$$ and $$^{2}$$

I think we should be able to prove the above relation or does change of basis (not changing commutator relations) only work for Hilbert spaces where $$T$$ (transformation matrices which are always unitary for orthonormal basis change for Hilbert space) on an operator behave as U()U^{\dagger} and not for a general change of basis like I tried
above for vector spaces which are .....um...non-Hilbert?

• That's not how a change of basis work on operators.
– fqq
Jul 30, 2021 at 18:39
• Yes @fqq is correct. Even for general basis change $A'=TAT^{-1}$ and not what you wrote as the relations. Jul 30, 2021 at 18:40
• Incidentally, it's not true that changes of basis are necessarily unitary in Hilbert spaces. They have no a priori relation to the scalar product structure.
– fqq
Jul 30, 2021 at 18:41
• Why? An operator is a square matrix and for any matrix, a change basis is simply a pre-multiplication with a transition matrix.
– Lost
Jul 30, 2021 at 18:41
• @Lost It's only one vote. One vote by itself means nothing. It happens all of the time. Jul 30, 2021 at 20:47

Summing over repeated indices, entries in the original commutator relation satisfy $$C_{ij}=A_{ik}B_{kj}-A_{kj}B_{ik}$$. The most general linear transformation of operators is $$A^\prime_{ij}=X_{ijmn}A_{mn}$$, and you're welcome to determine the condition on $$X$$ equivalent to $$X_{ijmn}(A_{mr}B_{rn}-A_{rn}B_{mr})=A_{ik}B_{kj}-A_{kj}B_{ik}$$. But as @fqq and @sslucifer note, if we want every vector to transform viz. $$v^\prime=Tv$$ we need $$TAv=A^\prime v^\prime=A^\prime Tv$$, so $$A^\prime=TAT^{-1}$$. (The condition $$T^{-1}=T^\dagger$$ preserves the inner product, but not all bases of interest are orthonormal.) This is the case $$X_{ijmn}=T_{im}(T^{-1})_{nj}$$, which you can verify works out. So depending on your perspective, you can see this transformation as multiplying $$A$$ by either one or two factors.
• @Lost this is a very good answer. The reason why $TAv=A'v'$ is because $Av$ is also a vector belonging to same vector space which is true for any Hilbert space. So the vector $Av$ must be transformed by the same relation. Jul 30, 2021 at 18:57