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The Hamiltonian for the quantum harmonic oscillator is

$$\hat{H}=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+\dfrac{1}{2}m\omega^2 x^2$$

and one can try to factorise it by writing down what later on will turn out to be ladder operators of the eigenspectrum

$$\begin{align}\hat{A}&=\sqrt{\dfrac{m\omega}{2\hbar}}\left(\hat{x}+\dfrac{i}{m\omega}\hat{p}\right)\\ \hat{A}^\dagger&=\sqrt{\dfrac{m\omega}{2\hbar}}\left(\hat{x}-\dfrac{i}{m\omega}\hat{p}\right)\end{align}$$

Now, in a problem class I'm supervising, the students were asked to "show that we can express the Hamiltonian $\hat{H}$ in terms of $\hat{A}^\dagger$ and $\hat{A}$", with the idea of obtaining the relation

$$\hat{H}=\hbar\omega\left(\hat{A}^\dagger\hat{A}+\dfrac{1}{2}\right)$$

The way the solution to this question is laid out is that the students should simply "guess" the combination $\hat{A}^\dagger\hat{A}$ is the right way to go, or get there by trial and error.

Question: what's the best/most intuitive way to explain why this is the case?

Writing $\hat{p}=-i\hbar\partial_x$, it's easy to justify taking some form of quadratic form of the operators, but why not e.g. just square them?

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  • $\begingroup$ if you understand German, p. 100 bottom + 101 in these lecture notes is quite useful: theorie.physik.uni-konstanz.de/burkard/sites/default/files/… $\endgroup$ – riddleculous Feb 16 '17 at 17:17
  • $\begingroup$ Square roots demand to be squared, and complex conjugates demand to be multiplied! $\endgroup$ – user12029 Feb 16 '17 at 17:20
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    $\begingroup$ It's clearer once you write the partials with respect to $x$ in terms of $\hat{p}$. Then you just harken back to algebra and factoring polynomials where $(a^2 - b^2)$ was a special enough case to be worth memorizing. $\endgroup$ – dmckee Feb 16 '17 at 17:24
  • $\begingroup$ @dmckee Yeah, I can see that - and it's how you get your operators in the first place. But going the other way round, is there any intuition as to why we take $\hat{A}^\dagger\hat{A}$ instead of $\hat{A}^2$, apart from checking that we do get rid of undesirable cross-terms? $\endgroup$ – Demosthene Feb 16 '17 at 17:32
  • $\begingroup$ Solve for $x$ and $p$ in terms of $A$ and $A^\dagger$. Then, substitute this into the Hamiltonian and simplify. $\endgroup$ – Prahar Feb 17 '17 at 0:34
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  • We know the explicit form of $A$ and $A^\dagger$ in terms of $p$ and $x$.
  • We know the expression of $H$ in terms of $p$ and $x$.

So just express $p$ and $x$ as a function of $A$ and $A^\dagger$, then plug the result in the formula for $H$. To do that, simply find $A + A^\dagger$ and $A - A^\dagger$, the rest will easily follow.

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How about this approach:

  • We know that position measurements $\hat{x}\left|n\right\rangle$ yield real numbers;
  • We know that momentum measurements $\hat{p}\left|n\right\rangle$ yield real numbers;
  • We know that energy measurements $\hat{H}\left|n\right\rangle$ are supposed to yield real numbers as well.

Given these factors, we note that if we wish to construct the operator $\hat{H}$ from some operator $\hat{A} \sim \hat{x}+i\hat{p}$, then we would somehow need to get rid of the $i$ to get real values for the energies. The obvious way to do that, would be to multiply $\hat{A}$ by its Hermitian conjugate to eliminate the $i$, and then seeing what we can do with $\hat{A}\hat{A}^\dagger$.

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One notes that $$ H=\frac{p^2}{2m}+\frac{1}{2}kx^2 = \frac{1}{2}k \left(x+\frac{ip}{m\omega}\right)\left(x-\frac{ip}{m\omega}\right)\, , $$ which suggests the form of the creation and destruction operators, up to appropriate constants.

In addition, the classical mechanical equations of motion for $x$ and $p$ for a harmonic oscillator are $$ \dot{x}=\frac{p}{m}\, ,\qquad \dot p=-kx\, . $$ Written in matrix form: $$ \frac{d}{dt}\left(\begin{array}{c} x \\ p\end{array}\right)= U\left(\begin{array}{c} x \\ p\end{array}\right)\, ,\qquad U=\left(\begin{array}{cc} 0 & 1/m \\ -k & 0 \end{array}\right)\, . $$ The matrix $U$ thus couples the evolution of $x$ and $p$. One can search for new variables $X$ and $P$ so that the evolutions of these are decoupled; this amounts to finding the eigenvectors of $U$.

It's an easy job to show that the eigenvalues of $U$ are $\pm i\omega$, and the eigenvectors $$ X={\cal A}\left(x+\frac{ip}{m\omega}\right)\, ,\qquad P={\cal B}\left(x-\frac{ip}{m\omega}\right)\, , \quad \omega^2=k/m\, , $$ which are proportional to the destruction and creation operators, respectively.

The first method is at the root of the Infeld-Hull factorization method, which in turn is closely related to superpotentials in supersymmetric quantum mechanics.

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You can make them look for the energy of every Fock state $ |n> $ and find out that it's $\hbar\omega(n+\frac{1}{2}) $ (oscillator's energy is quantized) and you can also see that $N=a^\dagger a$ is the number operator. $N|n>=n|n>$

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  • $\begingroup$ That would make sense. Unfortunately, I'm only the demonstrator, not the lecturer, and I have no say as to what goes in the problem sheets... So the whole thing is set up only in terms of explicit wavefunctions and the expressions for $\hat{A}$ I gave. $\endgroup$ – Demosthene Feb 16 '17 at 17:38
  • $\begingroup$ In fact, this is question 2, and question 1 is showing $[\hat{A},\hat{A}^\dagger]=1$ - I guess just to make sure students see they get consistent answers even if they choose $\hat{H}=\hbar\omega\left(\hat{A}\hat{A}^\dagger-\frac{1}{2}\right).$ $\endgroup$ – Demosthene Feb 16 '17 at 17:41
  • $\begingroup$ you still can put up the problem differently in the blackboard, that's what I would do. $\endgroup$ – Ismasou Feb 16 '17 at 17:49

protected by Qmechanic Feb 16 '17 at 20:25

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