Daily Archives: Fri, 3 February 2017

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)