# Linear Harmonic Oscillator

1. Jul 9, 2010

### Nemanja989

Hello there,

Can anyone help me, I am struggling with solving LHO in two dimension,but in the polar coordinates.
I transfer laplacian into polar from decart coordinates, write Ψ=ΦR, and do fourier separation method for solving differential equation. But I do not know how to solve differential equation on R.

I will write to where I came tomorrow, or in a few hours.

Thanks!!!

2. Jul 9, 2010

### Jivesh

Why trouble yourself by going into polar co-ordinates? Just look for solution of the form Ψ = X(x)Y(y), for the Schrodinger equation in cartesian co-ordinates and you'll get two equations for a Linear Harmonic Oscillator. Surely, for the case of two one dimensional equation, you know the solution, multiply them, and add there energy levels to get the answer. I think the lowest energy state would have the energy of (hbar)*w, where w is the natural frequency of oscillation and the wavefunction is a Gaussian in radial co-ordinate r.

3. Jul 9, 2010

### Dickfore

In this problem, both $\phi$ and $z$ are cyclic coordinates (they do not enter explicitly in the Schroedinger equation), so the corresponding conjugate momenta $p_{\phi} \equiv l_{z}$ and $p_{z}$ commute with the Hamiltonian. Therefore, the stationary states of the system are of the form:

$$\psi_{m, p_{z}}(\rho, \phi, z) = \frac{1}{(2 \pi \hbar)^{1/2}} \, e^{\frac{i}{\hbar} \, p_{z} \, z} \, \frac{1}{\sqrt{2 \, \pi}} \, e^{i \, m \, \phi} \, R_{m}(\rho), \ m = \ldots, -1, 0, 1, \ldots$$

Substitute this into the Schroedinger equation and you will get the following oridnary differential equation:

$$-\frac{\hbar^{2}}{2 \, \mu} \, \left[ \frac{1}{\rho} \, \frac{d}{d \rho}\left(\rho \, \frac{dR_{m}(\rho)}{d \rho}\right) - \frac{m^{2}}{\rho^{2}} \, R_{m}(\rho) - \frac{p^{2}_{z}}{\hbar^{2}} \, R_{m}(\rho) \right] + \frac{\mu \, \omega^{2} \, \rho^{2}}{2} \, R_{m}(\rho) = E \, R_{m}(\rho)$$

The normalization condition is:

$$\int_{0}^{\infy}{\rho \, R^{2}_{m}(\rho) \, d\rho} = 1$$

Introduce a dimensionless argument:

$$\rho = a x, R_{m}(\rho) = \frac{1}{a} \, y_{m}(x)$$

with:

$$a = \left(\frac{\hbar}{\mu \, \omega}\right)^{\frac{1}{2}}$$

and:

$$\epsilon = \frac{2}{\hbar \, \omega} \, \left( E - \frac{p^{2}_{z}}{2 \, \mu} \right)$$

you will get the following equation:

$$y''_{m} + \frac{1}{x} \, y'_{m}(x) + \left(\epsilon - x^{2} - \frac{m^{2}}{x^{2}}\right) \, y_{m}(x) = 0$$

with the normalization condition:

$$\int_{0}^{\infty}{x \, y^{2}_{m}(x) \, dx} = 1$$

Usually, we want to get rid of the first derivative in the differential equation. We can achieve this by the substitution:

$$y_{m}(x) = \frac{z_{m}(x)}{\sqrt{x}}$$

A direct substitution should convince you that the equation satisfied by $z_{m}(x)$ is:

$$z''_{m} + \left(\epsilon - x^{2} - \frac{m^{2} - 1/4}{x^{2}}\right) \, z_{m} = 0$$

$$\int_{0}^{\infty}{z^{2}_{m}(x) \, dx} = 1$$

The asymptotic behavior for $z_{m}(x)$ for both large and small values of $x$ is obtained by keeping the dominant terms in the equation for the relevant region:

$$z''_{m 0} - \frac{m^{2} - 1/4}{x^{2}} \, z_{m 0} = 0, \ x \rightarrow 0$$

This is an Euler equation and has solutions of the form $z_{m 0} \tilde x^{k}$. Substituting, we get an algebraic equation for $k$:

$$k^{2} - k - \left(m^{2} - \frac{1}{4}\right) = 0$$

the solutions of which are:

$$k_{1/2} = \frac{1 \pm 2 \, |m|}{2}$$

We take:

$$z_{m 0} \sim x^{|m| + \frac{1}{2}}, \ x \rightarrow 0$$

For large x, we have:

$$z''_{m \infty} - x^{2} \, z_{m \infty} = 0$$

which is of the same form as in the one dimensional case, so we have:

$$z_{m \infty} \sim e^{-\frac{x^{2}}{2}}, \ x \rightarrow \infty$$

Finally, we capture the asymptotic behavior by writing:

$$z_{m}(x) = x^{|m| + 1/2} \, e^{-x^{2}/2} \, v_{m}(x)$$

After some differentiation and algebraic manipulation, you should get the following equation for $v_{m}(x)$:

$$v''_{m} + \left(\frac{2 |m| + 1}{x} - 2 x \right) \, v'_{m} + \lambda \, v_{m} = 0$$

where

$$\lambda = \epsilon - 2 |m| - 1$$

Using a change of variables in the argument:

$$x = a \, t^{\alpha}$$

you will get:

$$t \, \ddot{v}_{m} + \left[ 1 + 2 \, \alpha \, |m| - 2 a^{2} \, \alpha \, t^{2 \, \alpha} \right] \, \dot{v}_{m} + \lambda \, (a \alpha)^{2} t^{2 \, \alpha - 1} = 0$$

Compare this with the differential equation for the Kummer confluent hypergeometric equation $_{1}F_{1}(a; c; t)$:

$$t \ddot{y} + (c - t) \, \dot{y} - a \, y = 0$$

we see that we should have:

$$2 \, \alpha = 1, \; 2 \, \alpha \, a^{2} = 1 \Rightarrow \alpha = \frac{1}{2}, \; a = 1 \Rightarrow x = t^{\frac{1}{2}} \Leftrightarrow t = x^{2}$$

So, we can write:

$$v_{m}(x) = _{1}F_{1}(-\frac{\lambda}{4}, |m| + 1, x^{2})$$

Collecting everything back, the radial wavefunction is:

$$R_{n m}(a \, x) = C_{n, m} \, x^{|m|} \, e^{-\frac{x^{2}}{2}} \, _{1}F_{1}\left( -n, |m| + 1, x^{2} \right), n \in \mathbb{N}_{0}, m \in \mathbb{Z}$$

where $C_{n, m}$ is a normalization constant. The energy eigenvalues associated with motion in the plane of the oscillator are:

$$E_{n, m, p_{z}} - \frac{p^{2}_{z}}{2 \, \mu} = E'_{n, m} = \hbar \, \omega \, \left( |m| + 2 \, n + \frac{1}{2} \right)$$

4. Jul 10, 2010

### Nemanja989

I have attached a pdf file into this post.

Thank you for responding!

You asked me why I do it in polar coordinates, when its much easier in decarte. Well I simply want to practice on my own, I want to solve one problem in a few ways. I have just finished basic cousre of quantum mechanics, but we havent done LHO in polar coordinates.

File size:
386.7 KB
Views:
173