# Variance of a harmonic motion particle

1. Feb 28, 2009

### TFM

1. The problem statement, all variables and given/known data

Consider a particle in the harmonic potential $$V(x)= \frac{m\omega^2 x^2}{2}$$. Its lowest energy eigenvalue is $$E_0=\hbar \omega/2$$ and the eigenfunction associated with this energy, ie the ground-state wave function, is

$$\phi_0(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{(m\omega/2\hbar)x^2}$$

a)

Determine the limits of the classical motion in this potential (the “classical turning points”), ie the smallest and the largest values of x that a classical particle can reach if it has the total energy E0.

b)

Assume that the wave function of the particle is the stationary state $$\phi_0(x) = e^{-iE_0t/\hbar}$$. Determine the probability (in the form of an integral) of finding the particle outside the region where classical motion can occur. By making an appropriate change of variable in the integral you obtain, show that the answer is independent of m, $$\omega$$, and $$\hbar$$.

c)

Calculate the variance $$\Delta \hat{x}$$ in the ground-state of the system and compare it to the limits of the classical motion.

useful integral:

$$\int^{\infty}_{-\infty} z^2 e^{-\alpha z^2} dz = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}$$

2. Relevant equations

Variance = $$\sqrt{<x^2> - <x>^2}$$

$$<x^2> = \int \phi^* x^2 \phi$$

$$<x> = \int \phi^* x \phi$$

3. The attempt at a solution

I have done all but the location of reflection for (a), and I have done (b), but I am slightly stuck for (c)

I have used the formulas given above, with:

$$\phi_0(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{(m\omega/2\hbar)x^2}$$

$$\phi_0(x)^* = \left(\frac{m^*\omega^*}{\pi \hbar^*}\right)^{1/4}e^{-(m\omega/2\hbar)x^2}$$

however when I put them in the variance formula, because it is e^x x e^-x, they cancel,so I am just left with:

$$<x> = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}\left(\frac{m^*\omega^*}{\pi \hbar^*}\right)^{1/4}x$$

and

$$<x^2> = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}\left(\frac{m^*\omega^*}{\pi \hbar^*}\right)^{1/4}x^2$$

This doesn't seem to coincide with the useful integer given...?

Have I gone wrong somewhere?

Many Thanks,

TFM

Last edited: Feb 28, 2009
2. Feb 28, 2009

### malawi_glenn

first of all, the wavefunction does not seem to be correct.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.html

$$\phi_0(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{(m\omega/2\hbar)x^2}$$
is wrong, it should be;
$$\phi_0(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{-(m\omega/2\hbar)x^2}$$

And now taking the complex conjugate of this one.. can you please tell me what the complex conjugate of a real number is? Then can you tell me if the ground state function is a real function?! ;-)

3. Feb 28, 2009

### TFM

The complex conjugate of a real number is just the real number.

there is no i's, so it is real...

this means

$$\phi_0(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{-(m\omega/2\hbar)x^2}$$

and

$$\phi_0(x)^* = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{-(m\omega/2\hbar)x^2}$$

so:

$$<x> = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{-(m\omega/2\hbar)x^2}\left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{-(m\omega/2\hbar)x^2} x$$

giving:

$$<x> = \int^{\infty}_{infty}\left(\frac{m\omega}{\pi \hbar}\right)^{1/2}e^{-2(m\omega/2\hbar)x^2} x dx$$

this looks lot better, but how would you use the integral

$$\int^{\infty}_{-\infty} z^2 e^{-\alpha z^2} dz = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}$$

for this one to solve? Would you use a substitiution?

4. Feb 28, 2009

### malawi_glenn

no for this integral you only need to do x^2 = y, and take y as integration variable.

Haven't you done calculus? :S

5. Feb 28, 2009

### TFM

Okay, so

$$<x> = \int^{\infty}_{infty}\left(\frac{m\omega}{\pi \hbar}\right)^{1/2}e^{-2(m\omega/2\hbar)x^2} x dx$$

take constants outside of integral

$$<x> = \left(\frac{m\omega}{\pi \hbar}\right)^{1/2}\int^{\infty}_{infty}xe^{-2(m\omega/2\hbar)x^2} dx$$

make

2(m\omega/2\hbar a constant, \alpha

$$<x> = \left(\frac{m\omega}{\pi \hbar}\right)^{1/2}\int^{\infty}_{infty}xe^{-\alpha x^2} dx$$

$$x^2 = y, x = \sqrt{y}$$

$$<x> = \left(\frac{m\omega}{\pi \hbar}\right)^{1/2}\int^{\infty}_{infty}\sqrt{y}e^{-\alpha y} dy$$

use integration by parts:

$$\int^{\infty}_{infty}\sqrt{y}e^{-\alpha y} dy$$

f(x) g'(x) = f(x)g(x) - \int{f'(x) g(x)}

f(x) = e^-ax
g'(x) = y

f'(x) = -(1/a)e^-ax
g(x) = 1

$$f(x) g'(x) = e^{-ax}1 - \int{-(1/a)e^-ax 1}$$

$$f(x) g'(x) = e^{-ax} - \int{-(1/a)e^-ax}$$

$$f(x) g'(x) = e^{-ax} - e^-ax}$$

$$f(x) g'(x) = e^{-ax} - e^-ax} = 0$$

Does this look right?

6. Mar 1, 2009

### malawi_glenn

no!

if x^2 = y, then xdx = (1/2)dy !!!!

Now I can give you one more approach, what is the integral over an odd function over an even interval? (can you use that in this situation?)

7. Mar 1, 2009

### TFM

what is the integral over an odd function over an even interval? (can you use that in this situation?)

0

Also,

$$= \left(\frac{m\omega}{\pi \hbar}\right)^{1/2}\int^{\infty}_{infty}xe^{-\alpha x^2} dx$$

using the substitution, gives:

$$\int^{\infty}_{infty}e^{-\alpha y} \frac{1}{2}dy$$

which is:

$$\frac{1}{2}\left[-\frac{1}{\alpha}e^{-\alpha y}\right]^{\infty}_{infty}$$

which is 0 from infinty to -infinty

Last edited: Mar 1, 2009
8. Mar 1, 2009

### malawi_glenn

yes, good!
but you can also see that it is 0 from the fact that xe^(-x^2) is an odd function.

Ok, lets do the x^2e^(-x^2) integral using that integral relation which was given to you.

9. Mar 1, 2009

### TFM

Okay, so:

$$<x^2> = \int \phi^* x^2 \phi$$

insert phi and phi*:

$$<x^2> = \int \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{(m\omega/2\hbar)x^2} x^2 \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{(m\omega/2\hbar)x^2}$$

giving:

$$<x^2> = \int \left(\frac{m\omega}{\pi \hbar}\right)^{1/2}e^{-(m\omega/\hbar)x^2} x^2$$

take out constants:

$$<x^2> = \left(\frac{m\omega}{\pi \hbar}\right)^{1/2} \int^{\infty}_{-\infty} e^{-(m\omega/\hbar)x^2} x^2$$

use:

$$\alpha = \frac{m\omega}{\hbar}$$

$$<x^2> = \left(\frac{\alpha}{\pi} \right)^{1/2} \int^{\infty}_{-\infty} e^{-\alpha x^2} x^2$$

use given integral:

$$<x^2> = \left(\frac{\alpha}{\pi} \right)^{1/2} \left[ \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}\right]$$

insert alpha:

$$<x^2> = \left(\frac{m\omega}{\hbar\pi} \right)^{1/2} \left[ \frac{1}{2}\sqrt{\frac{\pi}{\frac{m\omega}{\hbar}}}\right]$$

okay so far?

10. Mar 1, 2009

### malawi_glenn

no check the identity again

$$= \left(\frac{\alpha}{\pi} \right)^{1/2} \left[ \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}\right]$$

is wrong, it should be:

$$= \left(\frac{\alpha}{\pi} \right)^{1/2} \left[ \frac{1}{2}\sqrt{\frac{\pi}{\alpha ^3}}\right]$$

Slow down, don't hurry

11. Mar 1, 2009

### TFM

Okay so:

$$= \left(\frac{m\omega}{\hbar\pi} \right)^{1/2} \left[ \frac{1}{2}\sqrt{\frac{\pi}{\left(\frac{m\omega}{\hbar}\right)^{3}} }\right]$$

12. Mar 1, 2009

### malawi_glenn

simplify

13. Mar 1, 2009

### TFM

Okay,

$$= \left(\frac{m\omega}{\hbar\pi} \right)^{1/2} \left[ \frac{1}{2}\sqrt{\frac{\pi}{\left(\frac{m\omega}{\ hbar}\right)^{3}} }\right]$$

$$= \frac{1}{2}\sqrt{\frac{\frac{m\omega}{\hbar\pi}\pi}{\left(\frac{m\omega}{\hbar}\right)^{3}}$$

$$= \frac{1}{2}\sqrt{\frac{\frac{m\omega}{\hbar}}{\left(\frac{m\omega}{\hbar}\right)^{3}}$$

$$= \frac{1}{2}\sqrt{\frac{1}{\left(\frac{m\omega}{\hbar}\right)^{2}}$$

Would this go to:

$$= \frac{1}{2}\sqrt{\left(\frac{\hbar}{m\omega}\right)^{2}$$

?

14. Mar 1, 2009

### malawi_glenn

what is square root of quadratic?

15. Mar 1, 2009

### TFM

well x^2 square rooted is, so:

$$\frac{1}{2}\sqrt{\left(\frac{\hbar}{m\omega}\right )^{2}$$

can cancel to:

$$\frac{\hbar}{2m\omega}$$

16. Mar 1, 2009

### TFM

Okay, so now, if I put this into the variance:

$$\Delta \bar{x} = \sqrt{<x^2> - <x>^2}$$

$$\Delta \bar{x} = \sqrt{\frac{\hbar}{2m\omega} - 0^2}$$

$$\Delta \bar{x} = \sqrt{\frac{\hbar}{2m\omega}}$$

Which is very similar to the limits of the classical motion, which were

$$\sqrt{\frac{\hbar}{m\omega}}$$