# A unitary transformation induces a change $\delta\alpha$ on an operator that commutes with a complete set. Why must $\delta\alpha\propto 1$?

I'm reading Schwinger's 1951 paper "The Theory of Quantized Fields I". He described $$\alpha$$ as a complete set of commuting Hermitian operators, and considers a unitary transformation of them, defined as

$$\bar{\alpha} = U\alpha U^{-1}.$$

He then considers the infinitesimal transformation, i.e.

$$U = 1-(i/\hbar F), \quad F^\dagger = F,$$

which gives

$$\bar{\alpha} = \alpha - \delta \alpha, \quad i\hbar\delta\alpha = [\alpha,F].$$

So far so good. He then says

If the system is such that it is possible to obtain operators $$\delta\alpha$$ that commute with the complete set $$\alpha$$, one can treat the $$\delta\alpha$$ as arbitrary, infinitesimal numbers.

I've spent a while struggling with this line, and I just can't work out where it comes from. It simply isn't the case that an operator that commutes with the complete set of commuting operators must be proportional to the identity operator - any algebraic function of operators in the set itself has that property, after all. So how does he arrive at the conclusion that $$\delta \alpha$$ are "numbers"? I feel I must be missing something obvious.

if $$\delta \alpha$$ commute with your basis, then you can always write any product of operators $$\alpha_0 .. \delta \alpha .. \alpha_N$$ as $$\alpha_0 .. \alpha_N \delta \alpha$$.
Since $$\alpha_i$$ commute, $$\psi$$ can be written in a diagonal basis where $$\alpha_i \psi_k = a_{i,k} \psi_k$$ and $$a_{i,k}$$ is the eigenvalue. $$\delta \alpha$$ is also diagonal in this basis since it commutes with $$\alpha$$, therefore $$\delta \alpha ~ \psi_i = \delta a~ \psi_i$$ where $$\delta a$$ is a number.
Basically these conditions mean we can always work in a basis where $$\delta \alpha$$ is diagonal, and so we can always immediately apply it to states and treat it as a number. Because $$\delta a$$ is an infinitesimal, Schwinger says it's okay to treat the eigenvalue for $$\delta \alpha$$ as the same across the basis.
• Great, thank you, I think I get it - Schwinger meant you can "treat $\delta \alpha$ like numbers" in a somewhat more restricted sense than I took him to mean. That would be sufficient for the transformations to be uniquely labelled by elements of the reals though (you could pick the eigenvalues of any eigenvector as the label), which would then obey Abelian group composition laws according to the addition of their number-labels. Which is quite a substantial sense in which they're "like numbers"! Commented Dec 11, 2020 at 22:07