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Angular momentum operators in spherical coordinates

Note: variables with a hat are operators (such as $ \hat x$).


$\displaystyle \hat L_x$ $\displaystyle :=$ $\displaystyle i \hbar \left( \sin \phi \frac{\partial }{\partial \theta} + \frac{\cos \phi}{\tan \theta} \frac{\partial }{\partial \phi} \right) \hat x$ (14)
$\displaystyle \hat L_y$ $\displaystyle :=$ $\displaystyle i \hbar \left( - \cos \phi \frac{\partial }{\partial \theta} + \frac{\sin \phi}{\tan \theta} \frac{\partial }{\partial \pi} \right) \hat y$ (15)
$\displaystyle \hat L_z := - i \hbar \frac{\partial }{\partial \phi} \hat z = L_z \hat z$     (16)

So,

$\displaystyle H = T + V(r)$     (17)
  $\displaystyle =$ $\displaystyle \frac{p^2}{2m} + V(r)$ (18)
  $\displaystyle =$ $\displaystyle - \frac{\hbar^2}{2m} \nabla^2 + V(r)$ (19)

We can expand the term $ \hbar^2 \nabla^2$:,

$\displaystyle \hbar^2 \nabla^2 = \frac1r \frac{\partial^2}{\partial r^2} r + \frac{\hat L \cdot \hat L}{r^2}  $ (20)

We can write out the entire $ \hat L$ operator, and use the fact that $ \hat x$, $ \hat y$, and $ \hat z$ are orthogonal to compute $ L^2$:
$\displaystyle \hat L$ $\displaystyle =$ $\displaystyle L_x \hat x + L_y \hat y + L_z \hat z$ (21)
$\displaystyle L^2 = \hat L \cdot \hat L$ $\displaystyle =$ $\displaystyle L_x^2 + L_y^2 + L_z^2$ (22)

The key point here is that

$\displaystyle [H, \hat L] = 0,$    for $\displaystyle V = V(r)$ (23)

So, $ H$ and $ \hat L$ commute. If two operators commute, then you can measure the physical quantities associated with those simultaneously.

If $ [H, \hat L] \neq 0$, then when you measure energy (putting it in a well-defined state), then angular momentum is put into an unknown state of energy - even if you had previously put angular momentum into a well-defined state by measuring it.

In any physical system, it is useful to find the maximal set of operators which commute with each other.

In order to compute the commutator, the first part of $ H$ only has radial terms, so it will commute. The computation breaks down to: Does $ [L^2, \hat L] = 0$?

Recall the following relation between the components of angular momentum:

$\displaystyle [L_i, L_j] = \epsilon _{ijk} i \hbar L_k,$ (24)

where

$\displaystyle i, j, k \in \{x, y, z\} $

% latex2html id marker 2416
$\displaystyle \epsilon _{ijk} = \begin{cases}+1, & \text{for cyclic order} \\ -1, & \text{for anti-cyclic order} \end{cases}. $

To illustrate cyclic/anti-cyclic order:
$\displaystyle \epsilon _{xyz}$ $\displaystyle =$ $\displaystyle 1$  
$\displaystyle \epsilon _{zyx}$ $\displaystyle =$ $\displaystyle -1$  
$\displaystyle \epsilon _{yzx}$ $\displaystyle =$ $\displaystyle 1$  
$\displaystyle \epsilon _{xzy}$ $\displaystyle =$ $\displaystyle -1$  

We are ready to perform the commutator computation:

$\displaystyle [L^2, \hat L] = [L^2, L_x] \hat x + [L^2, L_y] \hat y + [L^2, L_z] \hat z  $ (25)

Consider:

$\displaystyle [L^2, L_x] = [L_x^2 + L_y^2 + L_z^2, L_x] = [L_y^2 + L_z, L_x]$ (26)

(as $ [L_x, L_x] = 0$ because $ L_x^2 L_x = L_x L_x^2$). Then,
$\displaystyle [L^2_y, L_x]$ $\displaystyle =$ $\displaystyle L_y^2 L_x - L_x L_y^2$ (27)
  $\displaystyle =$ $\displaystyle L_y^2 L_x - L_y L_x L_y + L_y L_x L_y - L_x L_y^2$ (28)
  $\displaystyle =$ $\displaystyle L_y (L_y L_x - L_x L_y) + (L_y L_x - L_x L_y) L_y$ (29)
  $\displaystyle =$ $\displaystyle L_y[L_y, L_x] + [L_y, L_x] L_y$ (30)

From (27):

$\displaystyle [L_y^2, L_x] = - i \hbar (L_y L_z + L_z L_y)$ (31)

We now consider $ [L_z^2, L_x]$. We can follow the same derivation process to get:
$\displaystyle [L_z^2, L_x]$ $\displaystyle =$ $\displaystyle L_z[L_z, L_x] + [L_z, L_x] L_y$ (32)
  $\displaystyle =$ $\displaystyle i \hbar (L_z L_y + L_y L_z)$ (33)

Plugging (34) and (36) into (29), we get

$\displaystyle [L^2, L_x] = [L_y^2 + L_z, L_x] = - i \hbar (L_y L_z + L_z L_y) + i \hbar (L_z L_y + L_y L_z) = 0$ (34)

Similarly,

$\displaystyle [L^2, L_y] = [L^2, L_z] = 0$ (35)

Also, $ [V(r), \hat L] = 0$. If $ V$ was a function of $ \hat r$ and not $ r$, then these would not commute.

We know that the eigenstates of the angular momentum operators are

$\displaystyle L^2 Y_{\ell, m}(\theta, \phi) = \ell(\ell+1) \hbar^2 Y_{\ell, m} (\theta, \phi) $

and

$\displaystyle L_z Y_{\ell, m} (\theta, \phi) = m \hbar Y_{\ell, m}(\theta, \phi) $

Comment: The maximal set of commuting operators is $ H$, $ L^2$, and $ L_z$. We could have chosen $ L_x$ or $ L_y$ instead of $ L_z$; we choose $ L_z$ because it has the simplest form in spherical coordinates.

Because of this, if we make measurements of $ E, \ell,$ and $ m$, then we collapse the wave function entirely.

In describing atoms with one electron, the interaction with the nucleus only depends on the Coulumb potential, which is spherical symmetrical. Without loss of generality, the wave function for an electron in a potential may be written

$\displaystyle \psi_{E, \ell, m} (r, \theta, \phi) = R_{E, \ell}(r) Y_{\ell, m}(\theta, \phi)$ (36)

The set of all these states is ``complete" in the sense that any wave function can be written in the above basis.

Hence, $ H \psi(r, \theta, \phi) = E \psi(r, \theta, \phi)$ implies:

$\displaystyle \left\{ -\frac{\hbar^2}{2 m r} \frac{\partial }{\partial r} r + \frac{\ell(\ell+1) \hbar^2}{2mr^2} + V(r) \right\} R(r) = E R(r) $

$ R(r)$ does not depend on $ m$.

Friday, 29 August 2008

Possible times for makeup classes:

Discussing HW problem: The time-dependent Schrödinger equation is:

$\displaystyle H \psi = i \hbar \frac{\partial \psi}{\partial t} $

If $ \psi(x,t) = \phi(x) e^{-Et}$, then $ \psi(x,t)$ is a stationary state - Why?

We say it is a stationary state because:

$\displaystyle \vert\psi(x,t)\vert^2 = \vert\phi(x)\vert^2. $

Hence, the probability of finding the particle at a place is independent of time.

We know solutions of

$\displaystyle H \phi_n(x) = E_{n} \phi_n(x) $

are complete. Meaning that any function, $ f(x)$ in the domain $ 0 \leq x \leq L$ can be expanded in the $ \phi_n(x)$. So,

$\displaystyle f(x) = \sum_{n=1}^\infty c_n \phi_n(x) $

From the sum, the probability that we find energy state $ E_n$ where $ n=2$ is $ \vert c_2\vert^2$.

Hermitian operators have eigenfunctions $ \phi_n$ such that $ \left\langle{ \phi_n, \phi_{n'} }\right\rangle = \delta_{n,n'}$ Here, the inner product is:

$\displaystyle \int_{-\infty}^\infty \phi_n^* \phi_{n'}(x) dx $

Thus,

$\displaystyle \left\langle{ \phi_{n'}, f }\right\rangle = \sum_{n=1}^\infty c_n\left\langle{ \phi_{n'}, \phi_n }\right\rangle = c_{n'} $

What is the wave function for this problem for $ t > 0$? Thus,

$\displaystyle f(x,t) = \sum_{n=1}^\infty c_n \phi_n(x) e^{-i/\hbar E_n t}$

The linear momentum is

$\displaystyle \hat p_x = - i \hbar \frac{\partial }{\partial x} $

The expected value is

$\displaystyle \left\langle{ f(x,t) \vert \hat p_x \vert f(x,t) }\right\rangle = \chi(t) $

$\displaystyle = \int_0^L f^*(x,t) \hat p_x f(x,t) dx = \chi(t) $

For the last problem of the homework,

$\displaystyle \langle \hat L_x \rangle = \langle \phi \vert \hat L_x \vert \phi \rangle $

Note about the book: Equation 7-28 is not quite correct. The $ 4 \pi$ only comes in because the book has not normalized the angular part of the wave function. Book uses $ \Theta_{\ell m}(\theta)$ and $ \Phi_m(\phi)$ instead of $ Y_{\ell m} (\theta, \phi)$.

Schrödinger equation for an energy eigenstate of a particle in a central potential $ V(r)$, $ \psi(r, \theta, \phi) e^{-i E t}$ satisfies:

$\displaystyle H \psi e^{-iEt} = - i \hbar \frac{\partial }{\partial t} \left( \psi e^{-i Et}\right) = E \psi e^{-i E t} $

Thus $ H \psi = E \psi$ may be written

$\displaystyle \left\{ -\frac{\hbar^2}{2m} \frac1r \frac{d^2}{dr^2} (r \cdots) +...
...hat L^2}{2mr^2} + V(r) \right\} \psi(r, \theta, \phi) = E \psi(r, \theta, \phi)$ (37)

We know

$\displaystyle \hat L^2 Y_{\ell m}(\theta, \phi) = \ell (\ell+1) \hbar^2 Y_{\ell m}(\theta, \phi) $

where $ Y_{\ell m} (\theta, \phi)$ is a spherical harmonic and

$\displaystyle \hat L_z Y_{\ell m}(\theta, \phi) = m \hbar Y_{\ell, m}(\theta, \phi) $

So write $ \psi(r, \theta, \phi) = R_{n \ell}(r) Y_{\ell m} (\theta, \phi)$ and substitute into (40), then

$\displaystyle \left\{ -\frac{\hbar^2}{2m} \frac1r \frac{d^2}{dr^2} r + \frac{\ell(\ell+1)\hbar^2}{2mr^2} + V(r) \right\} R_{n\ell}(r) = E_n R_{n \ell}(r)$ (38)

where $ n$ labels the energy eigenstate and eigenvalue.

Normalization of an energy eigenstate $ \psi_{n\ell m} = R_{n\ell} (r) Y_{\ell m}(\theta, \phi)$:


$\displaystyle 1$ $\displaystyle =$ $\displaystyle \int_{0}^\infty \int_0^{2 \pi} \int_0^\pi \vert\psi_{n \ell m}^2\vert \sin \theta r^2 dr d\phi d\theta$ (39)
  $\displaystyle =$ $\displaystyle \left( \int_0^\infty \vert R_{n \ell}(r)\vert^2 r^2 dr \right) \left( \int \int \vert Y_{\ell m}\vert^2 \sin \theta d\theta d\phi \right)$ (40)

Therefore, since

$\displaystyle \left\langle{ Y_{\ell m} }\right\rangle {Y_{\ell m}} = 1, $

we have

$\displaystyle \int_0^\infty r^2 \vert R_{n \ell}(r)\vert^2 = 1$ (41)

The radial probability is then $ P_{n\ell}(r) = r^2 \vert R_{n \ell}(r)\vert^2$.

Monday, 1 September 2008 Labor Day, no class.

Wednesday, 3 September 2008


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Next: General Central Potential Problem Up: Central Field Problems Previous: Quantum Treatment   Contents
Brian Bockelman 2008-09-11