Recurrence relations of the Hermite polynomials

The recurrence relations of the Hermite polynomials describe how each polynomial in the sequence can be obtained from its predecessors.

Expanding eq18 and assuming $n$ is an even number,

${H_n(y)=(-1)^{\frac{n}{2}}\frac{n!}{(\frac{n}{2})!}\biggr\[1-\frac{2n}{2!}y^2+\frac{2^2n(n-2)}{4!}y^4-\frac{2^3n(n-2)(n-4)}{6!}y^6+\cdots\biggr\]\;\;\;\;\;\;\;\;26}$

${H_{n-1}(y)=(-1)^{\frac{n-2}{2}}\frac{2(n-1)!}{(\frac{n-2}{2})!}\biggr\[y-\frac{2(n-2)}{3!}y^3+\frac{2^2(n-2)(n-4)}{5!}y^5-\frac{2^3(n-2)(n-4)(n-6)}{7!}y^7+\cdots\biggr\]\;\;\;\;\;\;\;\;27}$

Taking the derivative of eq26, we have

${\frac{dH_n(y)}{dy}=(-1)^{\frac{n}{2}-1}\frac{2(n-1)!}{(\frac{n-2}{2})!}\biggr\[2ny-\frac{2^2n(n-2)}{3!}y^3+\frac{2^3n(n-2)(n-4)}{5!}y^5-\cdots\biggr\]\;\;\;\;\;\;\;\;28}$

Substituting eq27 in eq28,

$\frac{dH_n(y)}{dy}=2nH_{n-1}(y)\;\;\;\;\;\;\;\;29$

If we repeat the steps from eq26 through eq29 on the assumption that $n$ is an odd number, we too end up with eq29. Therefore, $n$ in eq29 represents any number. Taking the derivative of eq29 again,

$\frac{d^2H_n(y)}{dy^2}=2n\frac{dH_{n-1}(y)}{dy}=4n(n-1)H_{n-2}(y)\;\;\;\;\;\;\;\;30$
Substitute 29 and 30 in eq23, we have

$H_n(y)-2yH_{n-1}(y)+2(n-1)H_{n-2}(y)=0\;\;\;\;\;\;\;\;31$

Replacing the dummy index $n$ in eq31 with $n+1$ gives

$H_{n+1}(y)-2yH_{n}(y)+2nH_{n-1}(y)=0\;\;\;\;\;\;\;\;32$

Eq29 and eq32 are the recurrence relations of the Hermite polynomials.

Question

Given $H_0(y)=1$ , show that eq32 can be used to generate the Hermite polynomials.

Answer

Substituting $n=0$ in eq32, we have $H_1(y)=2yH_0(y)=2y$ . Substituting $n=1$ in eq32, we have $H_2(y)=4y^2-2$ . Repeating the logic, we can generate the rest of the polynomials.

Question

Using eq32, ${y=\sqrt{\frac{m\omega}{\hbar}}x}$ and eq46, show that ${\langle\psi_k\vert x\vert\psi_n\rangle=\sqrt{\frac{\hbar(n+1)}{2m\omega}}\delta_{k,n+1}+\sqrt{\frac{n\hbar}{2m\omega}}\delta_{k,n-1}}$ .

Answer

Substituting eq32 in eq46, we have

$x\psi_n(x)=\sqrt{\frac{\hbar(n+1)}{2m\omega}}\psi_{n+1}(x)+\sqrt{\frac{n\hbar}{2m\omega}}\psi_{n-1}(x)$

Multiplying the above equation by $\psi^*_k(x)$ and integrating over all space,

$\langle\psi_k\vert x\vert\psi_n\rangle=\sqrt{\frac{\hbar(n+1)}{2m\omega}}\delta_{k,n+1}+\sqrt{\frac{n\hbar}{2m\omega}}\delta_{k,n-1}\;\;\;\;\;\;\;\;32a$

Recurrence relations of the Hermite polynomials

Question

Answer

Question

Answer

Next article: Taylor series of single-variable and multi-variable functions

Previous article: Orthogonality of the wavefunctions of The quantum harmonic oscillator

Content page of vibrational spectroscopy

Content page of advanced chemistry

Main content page

Leave a Reply Cancel reply