# Why does $[\hat{n},\hat{\phi}] = i$ imply $e^{i \hat{\phi}} |n\rangle = |n+1\rangle$?

When reading in quantum mechnics books and papers, I often come across a statement about two 'conjugate' variables. E.g.

Let be $$\hat{n}$$ and $$\hat{\phi}$$ two variables satisfying $$[\hat{n}, \hat{\phi}] = i$$. Then we know that $$e^{i \hat{\phi}} |n\rangle = |n+1\rangle$$ holds.

I've seen this statement quite often, but I have never found a proof or explanation of this fact. Probably it is something very fundamental that everybody (except me) knows by heart or has at least learnt in his quantum mechanics course. Unfortunately, non of both applies to me, so I am left back confused where this is coming from. I also tried to google it, but as I don't know the name of this relation I do not have a striking name to search for.

I tried to derive it from the commutation relation applying the Hadamard-Lemma and/or the Baker-Campbell-Hausdorff formula, but after some tedious calculations, I didn't obtain anything useful.

For a while I just tried to accept this fact as something given, but on the long term this is very unsatisfying. Therefore, I hope some of you can explain to me what's behind this relation.

• Does variable mean by the operator in your case? Nov 28, 2020 at 10:39
• Yes, that's the reason for ' ' at 'variable'. I've often read in connection with this statement that they are called conjugate variables, so wanted to use this term to draw a link for those who are familiar with this topi. Furthermore, if they were scalars, the commutator would be 0. Nov 28, 2020 at 10:44

Let us compute

$$\hat{n} \, e^{i \hat{\phi}} | n \rangle = \hat{n} \sum_{k=0}^\infty \frac{(i \hat{\phi})^k}{k!} | n \rangle = \sum_{k=0}^\infty \frac{i^k \big( [ \hat{n},\hat{\phi}^k] + \hat{\phi}^k \hat{n} \big)}{k!} | n \rangle = \sum_{k=0}^\infty \frac{i^k \big( k \, \hat{\phi}^{k-1} [\hat{n},\hat{\phi}] + \hat{\phi}^k \hat{n} \big)}{k!} | n \rangle = \\ = \sum_{k=0}^\infty \frac{i^k \big( k \, \hat{\phi}^{k-1} i + \hat{\phi}^k \hat{n} \big)}{k!} | n \rangle = \left( i^2 \sum_{k=1}^\infty \frac{i^{k-1} k \hat{\phi}^{k-1} }{k!} + \sum_{k=0}^\infty \frac{i^k \hat{\phi}^k }{k!} \hat{n} \right) | n \rangle \\ = \left( - e^{i \hat{\phi}} + e^{i \hat{\phi}} \hat{n} \right)| n \rangle = (n-1) e^{i \hat{\phi}} | n \rangle$$

where we have used that $$[a,b^n] = n \, b^{n-1} [a,b]$$. This shows that $$e^{i \hat{\phi}} | n \rangle$$ is an eigenvector of $$\hat{n}$$ with eigenvalue $$n-1$$, i.e. it is $$| n-1 \rangle$$. This is your result up to a sign, because I think you have a typo in your identity.

• Thank you very much! Nov 28, 2020 at 12:21

I did the following proof (it might be useful but it's pretty large though):

$$\hat n|n\rangle=n|n\rangle$$ $$[\hat n,\hat \phi]=i \ \ \ (\mathrm{given})$$ $$[\hat n,e^{i\hat \phi}]=-e^{i\hat \phi}$$

From the identity (see the proof here):

$$[\hat n,e^{i\hat \phi}]|n\rangle=(\hat ne^{i\hat \phi}-e^{i\hat \phi} \hat n)|n\rangle=\hat ne^{i \hat \phi}|n\rangle-ne^{i\hat \phi}|n\rangle$$ $$\Rightarrow -e^{i\hat \phi}|n\rangle =\hat ne^{i \hat \phi}|n\rangle-ne^{i\hat \phi}|n\rangle$$ Rearranging terms: $$\hat n(e^{i\hat \phi}|n\rangle)=(n-1)(e^{i\hat \phi}|n\rangle)$$ comparing with $$\hat{n}|n-1\rangle=(n-1)|n-1\rangle$$ $$e^{i\hat \phi}|n\rangle=(n-1)|n-1\rangle$$

There might be a scalar factor which I have omitted but can be found with the standard method used during ladder operator.

• Thank you for your answer, too! Unfortunately, I can only mark one answer as solution. Nevertheless, thank you very much for your solution! Nov 28, 2020 at 12:22
• It's no problem at all :) I write half of the answer when the above one gets posted so I thought, I should finish it. Nov 28, 2020 at 13:21