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I am trying to calculate the number operator Heisenberg picture using Baker –Hausdorf lemma:

After my calculation I end up by this: $$\exp\left(\frac{i}{\hbar} \hat{H} t\right)\,\hat{n}\, \exp\left(-\frac{i}{\hbar} \hat{H} t\right)= \hat{n}+[\hat{n},\,[\hat{H},\,\hat{n}]]+[\hat{H},\hat{n}]$$

I am trying to get further but I am not sure about what should I replace $\hat{H}$ by?

The answer should be just $\hat{n}$ for Heisenberg picture.

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  • $\begingroup$ Please use MathJax for equations - it makes them so much easier to read. $\endgroup$ – WetSavannaAnimal Apr 15 '18 at 4:19
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A hint: Assuming you are talking about the harmonic oscillator Hamiltonian (which is the context wherein the number operator makes sense), the Hamiltonian and the number operator commute, since one is a version of the other that has been scaled and "displaced" by the identity operator, which commutes with everything:

$$\hat{H} = \hbar\,\omega\,\left(a^\dagger \,a + \frac{1}{2}\,\mathrm{id}\right)=\hbar\,\omega\,\left(a\,a^\dagger- \frac{1}{2}\,\mathrm{id}\right) = \hbar\,\omega\left(\hat{n}- \frac{1}{2}\,\mathrm{id}\right)$$

A couple of points: The lemma you are using is often called the Campbell Baker Hausdorff theorem, but that's not the accepted usage. The lemma you are using should read:

$$\exp(X)\,Y\,\exp(-X) = Y + \mathrm{ad}_X\,Y + \frac{1}{2!} \mathrm{ad}_X^2\,Y+ \frac{1}{3!} \mathrm{ad}_X^3\,Y+ \cdots$$

where $\mathrm{ad}_X\,Y \stackrel{def}{=}[X,\,Y]$ and is generally true for bounded operators $X$, or for ones for which $\mathrm{ad}_X^m\,Y$ vanishes for some finite power $m$; the latter case being the one which often arises in QM, so that there is only a finite number of terms. Of course, $\mathrm{ad}_X\,Y=0$ when $X$ commutes with $Y$.

So I am not sure you have applied it altogether correctly, but you're getting the right conclusion!

This one is sometimes called the "braiding relationship", for the way it "braids" together the adjoint representations of a Lie group and the adjoint representation of the group's algebra. But, strictly speaking, Campbell-Baker-Hausdorff is a somewhat different beast.

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