1
$\begingroup$

In Griffiths Introduction to Quantum Mechanics, when discussing ladder operators in Chapter 2, he write Schrodinger's equation as, $$ \frac{1}{2m}\left[\left(\frac{\hbar}{\mathrm i}\,\frac{\mathrm d}{\mathrm dx}\right)^2+\left(m\omega x\right)^2\right]\psi=E\psi\tag{2.40} $$ Then he says,

The idea is to factor the term in the square brackets. If these were numbers, it would be easy: $$u^2+v^2=(u-iv)(u+iv).$$ Here, however, it's not quite so simple, because $u$ and $v$ are operators, and operators do not, in general, commute. Still this does invite us to look at the expressions, $$ a_\pm=\frac{1}{\sqrt{2m}}\left(\frac{\hbar}{\mathrm i}\frac{\mathrm d}{\mathrm dx}\pm m\omega x\right).\tag{2.41} $$

What is behind this idea to factorize? Equation 2.40 doesn't seem like it could be solved by factorising the term in the brackets. After all, the R.H.S of 2.40 is not zero, then why would one factorise to solve?

$\endgroup$
  • $\begingroup$ Why does the RHS need to be zero to factorize something $\endgroup$ – Kyle Kanos Dec 3 '19 at 11:49
  • $\begingroup$ Also, since we have MathJax enabled on the site, I've replaced the image with the portion of the text that seemed most relevant. You can look at the history of your post if you want to add more. $\endgroup$ – Kyle Kanos Dec 3 '19 at 12:04
  • 1
    $\begingroup$ The goal is to set up the ladder operators of QM; you are not solving a quadratic equation here. Hence the distinguishing nature of $a_\pm$ is clear, the factorisation of the the Schrodinger's equation is necessary, and non-problematic. $\endgroup$ – Kevin Dec 3 '19 at 15:13
  • $\begingroup$ You must not haveremoved the picture because it contained something that said something like: we are going to solve this using either power series method or algebraic method. So the thought that came into my mind first had to be that the writer was trying to solve it using factorisation and then he encountered the idea of ladder operators after he factorised. $\endgroup$ – Sufyan Naeem Dec 4 '19 at 5:35
3
$\begingroup$

The basic idea behind the factorization is to replace a 2nd-order differential equation by a pair of first order ones. It was made popular in physics by the work of Hull and Infeld:

Infeld, Leopold, and T. E. Hull. "The factorization method." Reviews of modern Physics 23.1 (1951): 21

although in fact earlier examples, such as the factorization of the harmonic oscillator into creation and destruction operators, were known before the aforementioned paper. The review paper is apparently openly accessible if you follow the GoogleScholar link

The method is actually quite general and Infeld and Hull do a good job of showing how one can factorize some 2nd-order differential equations.

It is at the core of supersymmetric quantum mechanics, where one finds operators $\hat A^\dagger$ and $\hat A$ so that $\hat A^\dagger \hat A$ and $\hat A\hat A^\dagger$ give two different Hamiltonians, connected by a superpotential. Suppose

\begin{align} \hat H\psi(x)=\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V_-(x)\right)\psi_0(x)=0\, , \qquad V_-(x)=V(x)-E_0 \end{align}

One can easily show that the Hamiltonian can be rewritten in the form \begin{align} \hat H_-=\frac{\hbar^2}{2m}\left(\frac{d^2}{dx^2}+\frac{\psi_0^{''}(x)}{\psi_0(x)}\right)=\hat A^\dagger\hat A\, , \end{align} where \begin{align} \hat A=\frac{\hbar}{\sqrt{2m}}\left(\frac{d}{dx}-\frac{\psi^{'}_0(x)}{\psi_0(x)}\right)\, ,\qquad \hat A^\dagger=\frac{\hbar}{\sqrt{2m}}\left(-\frac{d}{dx}-\frac{\psi^{'}_0(x)}{\psi_0(x)}\right)\, . \end{align} The interesting "supersymmetric" bit is that $\hat A\hat A^\dagger$ is the Hamiltonian for a different potential.

More details can be found on this in the review paper by Fred Cooper et al (e-copy here):

Cooper, Fred, Avinash Khare, and Uday Sukhatme. "Supersymmetry and quantum mechanics." Physics Reports 251.5-6 (1995): 267-385.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.