Quantum Science Philippines

Angular Momentum Solution (Hydrogen Atom)

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

Solution:

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

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

Then

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

Solution:

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

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

Finally,

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

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Solution of Radial Function (Hydrogen Atom)

Derive $\Big[\frac{d^{2}}{d\rho^{2}} - 2\lambda_{k,l} \frac{d}{d\rho} + \Big(\frac{2}{\rho} - \frac{l(l+1)}{\rho^{2}}\Big)\Big]Y_{k,l}(\rho)=0$ .

Solution:

To solve the equation derived from problem (1), we look at the asymptotic behavior. First, let us assume a series of solution. But we must take into account as our $\rho\rightarrow\infty$ .

A solution of the form:

$U_{k,l}\sim e^{\lambda_{k,l}\rho}$ and $e^{-\lambda_{k,l}\rho}$

But $e^{\lambda_{k,l}\rho}$ can be discarded since if $\rho\rightarrow\infty$ , it blows up. Then, in general, the solution for the equation at any distance is:

$U_{k,l}\sim e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)$ .

Substituting gives

$\frac{d^{2}}{d\rho^{2}}\Big(e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)\Big) + \Big(\frac{2}{\rho}-\frac{l(l+1)}{\rho^{2}}\Big)e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho) - \lambda_{k,l}^{2}e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho) = 0$ .

Taking the first term:

$\frac{d^{2}}{d\rho^{2}}\Big(e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)\Big) = \lambda_{k,l}e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)+e^-\rho{\lambda_{k,l}}\frac{d}{d\rho}\big[Y_{k,l}(\rho)\big]$

$\frac{d^{2}}{d\rho^{2}}\Big(e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)\Big) = \lambda_{k,l}e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho) - 2\lambda_{k,l}e^-\rho{\lambda_{k,l}}\frac{dY_{k,l}(\rho)}{d\rho} + e^-\rho{\lambda_{k,l}}\frac{d^{2}Y_{k,l}(\rho)}{d\rho^{2}}$ .

Therefore,

$\Big[\frac{d^{2}}{d\rho^{2}} - 2\lambda_{k,l}\frac{d}{d\rho} + \Big(\frac{2}{\rho}-\frac{l(l+1)}{\rho^{2}}\Big)\Big] Y_{k,l}(\rho)=0$ .

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Orbital Angular Momentum Operator Lz in Spherical Coordinates

by Alvin P. Aballe, MSU-IIT

Orbital angular momentum, including the total and spin angular momentum, plays a significant part in both classical and quantum mechanics. In classical mechanics, this is one of the conserve quantities alongside energy and linear momentum. With this, one can easily deal systems undergoing rotations. With that in mind, this will also be significant in dealing with the structure of the atom.

Like any other operators in quantum mechanics (position and linear momentum operator), orbital angular momentum as an operator could be expressed in different coordinate systems. The conversion of orbital angular momentum from one coordinate system to another could be convenient and efficient depending on the geometry of the system. So, given a system of spherical geometry, it is convenient to use the spherical form of this operator.

In 3D cartesian coordinate system, $\overset{\rightharpoonup}{L}(x, y, z) = ( L_x \hat x, L_y \hat y, L_z \hat z)$ .

In this article, we will express the z-component from cartesian to spherical coordinates.

Solution:

In classical mechanics,

$\overset{\rightharpoonup} L = \overset{\rightharpoonup}r \times \overset{\rightharpoonup}p.$

By replacing $\overset{\rightharpoonup}L, \overset{\rightharpoonup}r$ , and $\overset{\rightharpoonup}p$ with their operator counterparts, we can obtain the quantum mechanical orbital angular momentum operator which is

$\overset{\rightharpoonup} L = \overset{\rightharpoonup}r \times \overset{\rightharpoonup}p$

where $\overset{\rightharpoonup}p = -i\hbar\overset{\rightharpoonup}\nabla$ .

So,

$\begin{array}{lll} \overset{\rightharpoonup}L & = & \overset{\rightharpoonup}r \times (-i\hbar\overset{\rightharpoonup}\nabla) \\ & = & -i\hbar (\overset{\rightharpoonup}r \times \overset{\rightharpoonup}\nabla) \end{array}$

$\Rightarrow \overset{\rightharpoonup}L = -i\hbar \begin{array}{ |ccc|} \hat r & \hat\theta & \hat\phi \\ r & 0 & 0\\ \frac{\partial}{\partial r} & \frac{1}{r}\frac{\partial}{\partial\theta} & \frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\end{array}$

$\begin{array}{lll}\Rightarrow\overset{\rightharpoonup}L & = & -i\hbar (0\hat r - \frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\hat\theta + \frac{\partial}{\partial\theta} \hat\phi)\\ & = & -i\hbar (\frac{\partial}{\partial\theta}\hat\phi - \frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\hat\theta). \end{array}$

Recall that

$\begin{array}{lll} \hat x & = & \sin\theta\cos\phi\hat r -\sin\phi\hat\phi + \cos\theta\cos\phi\hat\theta\\ \hat y & = & \sin\theta\sin\phi\hat r + \cos\phi\hat\phi + \cos\theta\sin\phi\hat\theta\\ \hat z & = & \cos\theta\hat r - \sin\theta \hat\theta.\end{array}$

Now, if we take the dot product of $\overset{\rightharpoonup}L$ and $\hat z$ or $\overset{\rightharpoonup}L \cdot \hat z$ , we will eventually obtain expression for $L_z.$

So,

$\begin{array}{lll} \overset{\rightharpoonup}L \cdot \hat z & = & -i\hbar (\frac{\partial}{\partial\theta}\hat\phi - \frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\hat \theta) \cdot (\cos\theta\hat r - \sin\theta \hat\theta)\\ & = & -i\hbar ((0)\frac{\partial}{\partial\theta} - \frac{(-\sin\theta)}{\sin\theta}\frac{\partial}{\partial\phi})\\ & = & -i\hbar \frac{\sin\theta}{\sin\theta}\frac{\partial}{\partial\phi}\\ L_z & = & -i\hbar\frac{\partial}{\partial\phi}. \end{array}$

And thus, we successfully expressed orbital angular momentum operator z-component in spherical coordinates.

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Orbital Angular Momentum Operator Ly in Spherical Coordinates

by Alvin P. Aballe, MSU-IIT

In 3D cartesian coordinate system, $\overset{\rightharpoonup}{L}(x, y, z) = ( L_x \hat x, L_y \hat y, L_z \hat z)$ .

In this article, we will express the y-component from cartesian to spherical coordinates.

Solution:

In classical mechanics,

$\overset{\rightharpoonup} L = \overset{\rightharpoonup}r \times \overset{\rightharpoonup}p.$

$\overset{\rightharpoonup} L = \overset{\rightharpoonup}r \times \overset{\rightharpoonup}p$

where $\overset{\rightharpoonup}p = -i\hbar\overset{\rightharpoonup}\nabla$ .

So,

Recall that

Now, if we take the dot product of $\overset{\rightharpoonup}L$ and $\hat y$ or $\overset{\rightharpoonup}L \cdot \hat y$ , we will eventually obtain expression for $L_y.$

So,

$\begin{array}{lll} \overset{\rightharpoonup}L \cdot \hat y & = & -i\hbar (\frac{\partial}{\partial\theta}\hat\phi - \frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\hat\theta) \cdot (\sin\theta\sin\phi\hat r + \cos\phi\hat\phi +\cos\theta\sin\phi \hat\theta)\\ & = & -i\hbar ((\cos\phi)\frac{\partial}{\partial\theta} - (\cos\theta\sin\phi)\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}) \\ & = & i\hbar (-\cos\phi\frac{\partial}{\partial\theta} + \frac{\cos\theta\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi})\\ L_y & = & i\hbar (-\cos\phi\frac{\partial}{\partial\theta} + \frac{\sin\phi}{\tan\theta}\frac{\partial}{\partial\phi}). \end{array}$

And thus, we successfully expressed orbital angular momentum operator y-component in spherical coordinates.

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Orbital Angular Momentum Operator Lx in Spherical Coordinates

by Alvin P. Aballe, MSU-IIT

In 3D cartesian coordinate system, $\overset{\rightharpoonup}{L}(x, y, z) = ( L_x \hat x, L_y \hat y, L_z \hat z)$ .

In this article, we will express the x-component from cartesian to spherical coordinates.

Solution:

In classical mechanics,

$\overset{\rightharpoonup} L = \overset{\rightharpoonup}r \times \overset{\rightharpoonup}p.$

$\overset{\rightharpoonup} L = \overset{\rightharpoonup}r \times \overset{\rightharpoonup}p$

where $\overset{\rightharpoonup}p = -i\hbar\overset{\rightharpoonup}\nabla$ .

So,

Recall that

Now, if we take the dot product of $\overset{\rightharpoonup}L$ and $\hat x$ or $\overset{\rightharpoonup}L \cdot \hat x$ , we will eventually obtain expression for $L_x.$

So,

$\begin{array}{lll} \overset{\rightharpoonup}L \cdot \hat x & = & -i\hbar (\frac{\partial}{\partial\theta}\hat\phi - \frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\hat\theta) \cdot (\sin\theta\cos\phi\hat r - \sin\phi\hat\phi + \cos\theta\cos\phi\hat\theta)\\ & = & -i\hbar ((-\sin\phi)\frac{\partial}{\partial\theta} - (\cos\theta\cos\phi)\frac{1}{\sin\theta}\frac{\partial}{\partial\phi})\\ & = & i\hbar (\sin\phi\frac{\partial}{\partial\theta} + \frac{\cos\theta\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi})\\ L_x & = & i\hbar (\sin\phi\frac{\partial}{\partial\theta} + \frac{\cos\phi}{\tan\theta}\frac{\partial}{\partial\phi}). \end{array}$