Angular Momentum Solution (Hydrogen Atom) | Quantum Science Philippines
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Angular Momentum Solution (Hydrogen Atom)

a. ) Show that \big[-l(l+1)+s(s+1)\big]C_{0}=0.


To obtain a solution to the equation derived in Problem (2), we assume a power series solution of the form:

Y_{k,l}=\rho ^{s}\sum_{q}^{\infty}C_{q}\rho ^{q}=\sum_{q}^{\infty}C_{q}\rho ^{s+q} .


\frac{d}{d\rho}Y_{k,l}=\sum_{q}^{\infty}(q+s)C_{q}\rho ^{s+q-1} and

\frac{d^{2}}{d\rho^{2}}Y_{k,l}=\sum_{q}^{\infty}(q+s)(q+s-1)C_{q}\rho ^{s+q-2}

Substituting this to the equation solved in (2), we have

\sum_{q}^{\infty}(q+s)(q+s-1)C_{q}\rho ^{s+q-2}-2\lambda_{k,l}\sum_{q}^{\infty}(q+s)C_{q}\rho ^{s+q-1} + 2\sum_{q}^{\infty}C_{q}\rho ^{s+q-1}-\sum_{q}^{\infty} l(l+1) C_{q}\rho ^{s+q-2} = 0


\big[s(s-1)-l(l+1)\big]C_{0}\rho ^{s-2} \big[s(s+1)C_{1}-2\lambda_{k,l} s C_{0} + 2C_{0}-l(l+1)C_{1}\big]\rho ^{s-1} + \sum_{q=2}^{\infty}(q+s)(q+s-1)C_{q}\rho ^{s+q-2}-\sum_{q=1}^{\infty}(q+s)\rho ^{s+q-1}C_{q} + \sum_{q=1}^{\infty}2C_{q}\rho ^{s+q-1}- \sum_{q=2}^{\infty}l(l+1)C_{q}\rho ^{s+q-2}=0

Since \rho is linearly independent from all the rest, then each coefficient is equal to zero. That is,

\big[s(s+1) - l(l+1)\big] C_{0}=0.


b.) Show that \big[q(q+2l+1)\big]C_{q}=2\big[(q+1)\lambda_{k,l}-1\big]C_{q-1}.


Since each coefficient is equal to zero, we find the coefficient of “\rho ^{q+s-2} from the equation in Prob. (3). That is,

\sum_{q}^{\infty}(q+s)(q+s-1)C_{q}\rho ^{s+q-2}-2\lambda_{k,l}\sum_{q}^{\infty}(q+s)C_{q}\rho ^{s+q-1} + 2\sum_{q}^{\infty}C_{q}\rho ^{s+q-1}-\sum_{q}^{\infty} l(l+1) C_{q}\rho ^{s+q-2} = 0

Changing q\rightarrow q-1:

\sum_{q=2}^{\infty}\bigg[(q+s)(q+s-1)C_{q}-2\lambda_{k,l}(q+s-1)C_{q-1} + 2C_{q-1} - l(l+1)C_{q}\bigg]\rho ^{q+s-2}

\bigg[(q+s)(q+s-1)-l(l+1)\bigg]C_{q} -\bigg[ 2\lambda_{k,l}(q+s-1) - 2\bigg] C_{q-1}=0.

Now, taking s=l+1, we have

\bigg[(q+l+1)(q+l)-l(l+1)\bigg]C_{q} = \bigg[ 2\lambda_{k,l}(q+1) - 2\bigg] C_{q-1}


\bigg[q(q+2l+1)\bigg]C_{q} = 2\bigg[ \lambda_{k,l}(q+1) - 1\bigg] C_{q-1}.

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