# Exponential of an operator plus a constant term

I am reading a book on operator and matrix representation. Most of the examples are on Physics and they mention many terms like 'commute', i.e. the order of application of two operators might be matter. Anyway, let say I have an operator $\hat{x}$ which is within the exponential $\exp(\hat{x})$, we know that this could be expanded in matrix form. If there is one more operator $\hat{y}$ such that $\exp(\hat{x} + \hat{y})$, generally, we don't have $$\exp(\hat{x} + \hat{y}) = \exp(\hat{x})\exp(\hat{y}).$$ I don't know why and what leads to above conclusion but I just read this from the book. My question is if instead adding the second operator, we add a number ($\phi$) to the first operator like $\exp(\hat{x} + \phi)$, is it ALWAYS true to have $$\exp(\hat{x} + \phi) = \exp(\hat{x})\exp(\phi) = \exp(\phi)\exp(\hat{x})$$ even $\hat{x}$ is an operator?

Thanks

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Yes, this is true whenever the operators $\hat x$ and $\hat y$ commute. – Marek Jan 17 '13 at 12:31
Thanks Marek. So if y is number instead of operator, it always commute with any operator, is that right? – user1285419 Jan 17 '13 at 20:59
What is denoted by a number really means a scalar operator in this contest, i.e., multiplication by said number. A scalar operator commutes with any linear operator T because T(cx) = cT(x) by the linearity of T. – user53153 Jan 18 '13 at 5:59