how to find the ladder operators for this hamiltonian: $$\widehat{H}=a\widehat{A}^2 + b\widehat{B}^2$$ where $a$ and $b$ are two real and positive constants.

And how to write the hamiltonian in function of the two ladder operators?

Actually the answer is:$$a_-=\frac{1}{\sqrt{2a}}\widehat{A}+i\frac{1}{\sqrt{2b}}\widehat{B}$$ and $$a_+=\frac{1}{\sqrt{2a}}\widehat{A}-i\frac{1}{\sqrt{2b}}\widehat{B}$$

And the condition on the commutator $\widehat{A}$ and $\widehat{B}$ for having: $[a_-,a_+]=\widehat{1}$, I found: $$[\widehat{A},\widehat{B}]=i\sqrt{ab}$$

but I couldn't reach them.

Thank you in advance.

  • $\begingroup$ You need to define the commutation relations of $\hat A,\hat B$ for ladder operators to make sense $\endgroup$ – user619894 Dec 31 '19 at 11:51
  • $\begingroup$ ok done @user721481 $\endgroup$ – walid Dec 31 '19 at 11:59
  • $\begingroup$ To get the commutation you had to compute $\hat A,\hat B$ in terms of $a_{+,-}$ so you are basically done. $\endgroup$ – user619894 Dec 31 '19 at 12:16
  • $\begingroup$ Cool thank you it was a good hint. I was trying to find how to factorize the expression of H to deduce the ladder operators? $\endgroup$ – walid Dec 31 '19 at 13:03
  • $\begingroup$ Does the question mark mean you are still having trouble? If so I will elaborate. $\endgroup$ – user619894 Dec 31 '19 at 13:18

Moving this to an answer as in is too long for a comment: Given a generic Hamiltonian $a\widehat{A}^2 + b\widehat{B}^2$ is not enough information to construct ladder operators. There is no reason to expect that the commutation relation of two conjugate linear combinations of $A,B$ is 1. So you need to be supplied with either the commutation relations of $[A,B]$ or the explicit construction of $A,B$ in terms if $a_{+},a_{-}$.

In the question you say that



So you can plug back in and get the Hamiltonian in terms of $a_{+}, a_{-} $ but the Hamiltonian will now contain mixed terms: $\alpha a_{+}^2+\beta a_{-}^2+\gamma a_{+} a_{-}+\delta a_{-} a_{+}$ for some parameters $\alpha,\beta,\gamma,\delta$.

on the other hand, if you are given $[\widehat{A},\widehat{B}]=i\sqrt{ab}$ , then the construction of a pair of conjugate operators that have a commutation of $1$ is also direct.

  • $\begingroup$ I see, actually, they don't ask me to find a+ and a- they are given to work with them. But I try to refind them from [A, B]=i√(ab), to do so I get stuck on how to write H as a product, like $x^2+y^2=(x+iy)(x−iy) $ $\endgroup$ – walid Dec 31 '19 at 15:03

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