Hermite polynomials – generation

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.16.

In the solution of the Schrödinger equation for the harmonic oscillator, we found that the wave function can be expressed as a power series:

\displaystyle  \psi(y)=e^{-y^{2}/2}\sum_{j=0}^{\infty}a_{j}y^{j}

where {y} was introduced as a shorthand variable:

\displaystyle  y\equiv\sqrt{\frac{m\omega}{\hbar}}x

and the coefficients {a_{j}} satisfy the recursion relation

\displaystyle  a_{j+2}=\frac{2j+1-\epsilon}{(j+1)(j+2)}a_{j} \ \ \ \ \ (1)


where {\epsilon} is another shorthand variable for the energy:

\displaystyle  \epsilon\equiv\frac{2E}{\hbar\omega}

By requiring the recursion relation to terminate at various values of {j} we can generate the polynomials for the various energy states, which turn out to be the Hermite polynomials.

For example, if we take the highest value of {j} to be 5, then we must have

\displaystyle   a_{7} \displaystyle  = \displaystyle  \frac{11-\epsilon}{42}a_{5}
\displaystyle  \displaystyle  = \displaystyle  0
\displaystyle  \epsilon \displaystyle  = \displaystyle  11
\displaystyle  E \displaystyle  = \displaystyle  \frac{11}{2}\hbar\omega

To get the coefficients, we can start with {a_{0}=0} (since all even terms must be zero if we want {a_{5}\ne0}) and {a_{1}=1}. Then we get

\displaystyle  a_{j+2}=\frac{2j-10}{\left(j+1\right)\left(j+2\right)}a_{j}

so {a_{3}=-\frac{4}{3}}, {a_{5}=\left(-\frac{1}{5}\right)\left(-\frac{4}{3}\right)=\frac{4}{15}}. The Hermite polynomial is

\displaystyle  H_{5}(x)=A_{5}\left[\frac{4}{15}x^{5}-\frac{4}{3}x^{3}+x\right]

where {A_{5}} is a constant that is set by convention to make the coefficients satisfy some specified rule.

If we require the coefficient of the highest power {n} to be {2^{n}} we can multiply this polynomial by 120 to get

\displaystyle  H_{5}(x)=32x^{5}-160x^{3}+120x

For {H_{6}}, all the odd terms are zero, and we require {a_{8}=0}, so we get

\displaystyle   \epsilon \displaystyle  = \displaystyle  13
\displaystyle  E \displaystyle  = \displaystyle  \frac{13}{2}\hbar\omega
\displaystyle  a_{j+2} \displaystyle  = \displaystyle  \frac{2j-12}{\left(j+1\right)\left(j+2\right)}a_{j}

Taking {a_{0}=1} and {a_{1}=0}, we get {a_{2}=-6}, {a_{4}=4}, {a_{6}=-\frac{8}{15}}. Requiring the coefficient of {a_{6}} to be {2^{6}=64} we get

\displaystyle  H_{6}(x)=64x^{6}-480x^{4}+720x^{2}-120

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