# Harmonic oscillator: momentum space functions and Hermite polynomial recursion relations from raising and lowering operators

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.5, Exercises 7.5.1 – 7.5.3.

Earlier, we found the position space energy eigenfunctions of the harmonic oscillator to be

 $\displaystyle \psi_{n}(y)$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2}\ \ \ \ \ (1)$ $\displaystyle \psi_{n}(x)$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (2)$

where ${y}$ in the first equation is shorthand for

$\displaystyle y=\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (3)$

It turns out that an alternative method for deriving these functions uses the lowering operator ${a}$. Shankar gives the derivation of ${\psi_{n}\left(x\right)}$ in his section 7.5, but we can use the same technique to derive the momentum space functions. We start with the ground state and use

$\displaystyle a\left|0\right\rangle =0 \ \ \ \ \ (4)$

In terms of ${X}$ and ${P}$, we have

$\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}X+i\frac{1}{\sqrt{2m\omega\hbar}}P \ \ \ \ \ (5)$

To find the momentum space functions, we need to express ${X}$ and ${P}$ in terms of ${p}$:

 $\displaystyle X$ $\displaystyle =$ $\displaystyle i\hbar\frac{d}{dp}\ \ \ \ \ (6)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle p \ \ \ \ \ (7)$

We thus have

$\displaystyle \left[i\hbar\sqrt{\frac{m\omega}{2\hbar}}\frac{d}{dp}+i\frac{1}{\sqrt{2m\omega\hbar}}p\right]\psi_{0}\left(p\right)=0 \ \ \ \ \ (8)$

If we define the auxiliary variable

$\displaystyle z\equiv\frac{p}{\sqrt{\hbar m\omega}} \ \ \ \ \ (9)$

we get

$\displaystyle \left(\frac{d}{dz}+z\right)\psi_{0}\left(z\right)=0 \ \ \ \ \ (10)$

This has the solution

$\displaystyle \psi_{0}\left(z\right)=Ae^{-z^{2}/2} \ \ \ \ \ (11)$

for some normalization constant ${A}$. Thus in terms of ${p}$ we have

$\displaystyle \psi_{0}\left(p\right)=Ae^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (12)$

Normalizing in the usual way, making use of the Gaussian integral, we have

 $\displaystyle \int_{-\infty}^{\infty}\psi_{0}^{2}\left(p\right)dp$ $\displaystyle =$ $\displaystyle A^{2}\int_{-\infty}^{\infty}e^{-p^{2}/\hbar m\omega}dp=1\ \ \ \ \ (13)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{1}{\left(\pi\hbar m\omega\right)^{1/4}} \ \ \ \ \ (14)$

This agrees with the earlier result which was obtained by solving a second-order differential equation.

We can also use ${a}$ and ${a^{\dagger}}$ to verify a couple of recursion relations for Hermite polynomials. Reverting back to position space we have

 $\displaystyle X$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (15)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle -i\hbar\frac{d}{dx} \ \ \ \ \ (16)$

so 5 becomes

$\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}x+\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (17)$

Also from 5 we have, since ${X}$ and ${P}$ are both hermitian operators

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{2\hbar}}X-i\frac{1}{\sqrt{2m\omega\hbar}}P\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{2\hbar}}x-\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (19)$

Defining

$\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (20)$

we have

 $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(y+\frac{d}{dy}\right)\ \ \ \ \ (21)$ $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(y-\frac{d}{dy}\right) \ \ \ \ \ (22)$

We also recall the normalization conditions on the raising and lowering operators:

 $\displaystyle a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (23)$ $\displaystyle a^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (24)$

Applying 23 to 1 we have, after cancelling common factors from each side:

 $\displaystyle \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2^{n}n!}}\left(y+\frac{d}{dy}\right)\left[H_{n}(y)e^{-y^{2}/2}\right]$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (25)$ $\displaystyle \frac{1}{2\sqrt{n}}\frac{1}{\sqrt{2^{n-1}\left(n-1\right)!}}e^{-y^{2}/2}\left[yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy}\right]$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (26)$ $\displaystyle yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy}$ $\displaystyle =$ $\displaystyle 2nH_{n-1}\left(y\right)\ \ \ \ \ (27)$ $\displaystyle H_{n}^{\prime}\left(y\right)$ $\displaystyle =$ $\displaystyle 2nH_{n-1}\left(y\right) \ \ \ \ \ (28)$

Another recursion relation for Hermite polynomials can be found as follows. We start with 22 to get

$\displaystyle a+a^{\dagger}=\sqrt{2}y \ \ \ \ \ (29)$

We now apply 23 and 24 to 1. We can cancel common factors, including ${e^{-y^{2}/2}}$, from both sides to get

 $\displaystyle \left(a+a^{\dagger}\right)\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{2}y\psi_{n}\ \ \ \ \ (30)$ $\displaystyle \frac{\sqrt{2}y}{\sqrt{2^{n}n!}}H_{n}(y)$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{\sqrt{n+1}}{\sqrt{2^{n+1}\left(n+1\right)!}}H_{n+1}(y)\ \ \ \ \ (31)$ $\displaystyle \frac{y}{\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n}(y)$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{1}{2\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n+1}(y)\ \ \ \ \ (32)$ $\displaystyle yH_{n}\left(y\right)$ $\displaystyle =$ $\displaystyle nH_{n-1}\left(y\right)+\frac{1}{2}H_{n+1}\left(y\right)\ \ \ \ \ (33)$ $\displaystyle H_{n+1}\left(y\right)$ $\displaystyle =$ $\displaystyle 2yH_{n}\left(y\right)-2nH_{n-1}\left(y\right) \ \ \ \ \ (34)$