Laddering

Then, if

$$H\vert n> = \epsilon_n\vert n>$$ = \epsilon_n\vert n> \end{displaymath}">

operating from the left with M⁺ one has

$$M^+ H\vert n> = \epsilon_n M^+\vert n>=\epsilon_n \vert n+1>$$ = \epsilon_n M^+\vert n>=\epsilon_n \vert n+1> \end{displaymath}">

and since [H,M⁺] = HM⁺ - M⁺H, using Equation 5 one has

$$HM^+\vert n>-[H,M^+] \vert n> =\epsilon_n \vert n+1>$$-[H,M^+] \vert n> =\epsilon_n \vert n+1> \end{displaymath}">

$$H\vert n+1>-[H,M^+] \vert n> = \epsilon_n \vert n+1>$$-[H,M^+] \vert n> = \epsilon_n \vert n+1> \end{displaymath}">

and then, after a tedious amount of algebra, one obtains

$$H\vert n+1> +\left [ 2n \left ( \frac{\pi}{L}\right )^2 - \f... ... ( \left (\frac{\pi}{L}\right )^2 (2n + 1) \right )\vert n+1>$$ +\left [ 2n \left ( \frac{\pi}{L}\right )^2 - \f... ... ( \left (\frac{\pi}{L}\right )^2 (2n + 1) \right )\vert n+1> \end{displaymath}">

From this we conclude that, if the elements in square brackets cancelled perfectly, M⁺|n+1> would be an eigenfunction, i.e., if

$$\left [ 2n \left ( \frac{\pi}{L}\right )^2 - \frac{4 m}{n \hbar^2} \epsilon_n \right ] = 0$$

i.e.,

$$\epsilon_n = \frac{n^2 \hbar^2 \pi^2}{2 m L^2}$$

then

$$H\vert n+1> = \left ( \epsilon_n + \frac{\hbar^2}{2m} \left ... ... (\frac{\pi}{L}\right )^2 (2n + 1) \right )\right )\vert n+1>$$ = \left ( \epsilon_n + \frac{\hbar^2}{2m} \left ... ... (\frac{\pi}{L}\right )^2 (2n + 1) \right )\right )\vert n+1> \end{displaymath}">

$$= \frac{\hbar^2 \pi^2}{2mL^2} \left ( n^2+ 2n + 1 \right )\v... ...eft ( (n + 1)^2 \right )\vert n+1> = \epsilon_{n+1} \vert n+1>$$ = \epsilon_{n+1} \vert n+1> \end{displaymath}">

This represents a deluge of results, since we have obtained the redundant result of the form of the eigenvalue and the form of the next (higher) eigenvalue. Generally, we have to work a bit harder in ladder operator solutions to quantum mechanical problems, than we have had to here.