I tried to solve the quantum harmonic oscillator via the operator method. After doing it and looking up the solution I noticed that for some reason the ladder operators got an additional factor of (i) / (-i) in them. Is this just a (weird) convention? In my calculations I don't seem to need it and since only a product appears in the Hamiltonian it does not matter if there is an (i)*(-i)=1 in there or not.

My calculations are made with pencil - the extra (i) & (-i) is in blue.

Did I make some mistake or is really just a convention?enter image description here


closed as off-topic by ACuriousMind, Kyle Kanos, Neuneck, DanielSank, Martin Aug 7 '15 at 13:06

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  • $\begingroup$ Please type all relevant information into the post. Posting pictures of calculation is discouraged. We much prefer you type in whatever you need. There are several reasons for this: 1) It avoids link rot, 2) it's easier to read, 3) you are much more likely to figure out what parts are actually relevant to your question, which leads to more focused questions and possibly you figuring it out yourself. $\endgroup$ – DanielSank Aug 5 '15 at 15:57
  • $\begingroup$ It actually is even a bit more extreme than @BillN suggested: you have a quadratic Hamiltonian $\alpha^2 x^2 + \beta^2 p^2$, which is a Euclidean-norm $r^2$ acting on "distances" $\alpha x$ and $\beta p$. There's no reason that you can't choose some rotated axes in this space, like $\sqrt{1/2}(\alpha x \pm \beta p)$, to be your "quadratures" for the harmonic oscillator. In the end one dimensionless axis is $\hat a^\dagger + \hat a$ and the other is $i\hat a^\dagger - i \hat a$, your Hamiltonian is $\hbar\omega\left(\hat a^\dagger \hat a + \frac 12\right)$, and everything else is up in the air. $\endgroup$ – CR Drost Aug 5 '15 at 16:04
  • $\begingroup$ I think "check-my-work" questions are not considered on topic here. $\endgroup$ – DanielSank Aug 6 '15 at 17:02
  • $\begingroup$ I'm voting to close this question as off-topic because it is a check-my-work question. Please see if you can isolate a specific conceptual question. $\endgroup$ – DanielSank Aug 6 '15 at 17:03

There really is not a unique way to define the ladder operators, especially with the constants in front. If you change them from what one book has, you might match another book. Those differences will show up in different normalization constants for "next" states that the operators create.

Eventually, you will use your operators to find the algebraic form of the eigenstates, and those eigenstates, when properly normalized, will look identical to every else's.

So a difference in a factor of $i$ when defining the ladder operators is not unusual. Some people even call them $b$ and $b^\dagger$!


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