# Harmonic oscillator: statistics

Required math: algebra, calculus (partial derivatives and integration by parts), complex numbers

Required physics: Schrödinger equation, probability density

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.9.

Suppose a particle is in the quantum state

$\displaystyle \Psi\left(x,t\right)=Ae^{-amx^{2}/\hbar}e^{-iat} \ \ \ \ \ (1)$

where ${A}$ is the normalization constant and ${a}$ is a constant with dimensions of 1/time. We can find ${A}$ from normalization:

 $\displaystyle \int_{-\infty}^{\infty}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\int_{-\infty}^{\infty}e^{-2amx^{2}/\hbar}dx\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\sqrt{\frac{\pi\hbar}{2ma}}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \left(\frac{2ma}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (5)$

The spatial component of the wave function is

$\displaystyle \psi\left(x\right)=\left(\frac{2ma}{\pi\hbar}\right)^{1/4}e^{-amx^{2}/\hbar} \ \ \ \ \ (6)$

and it must satisfy the time-independent Schrödinger equation in one dimension

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$ $\displaystyle =$ $\displaystyle E\psi(x) \ \ \ \ \ (7)$

The energy ${E}$ can be found from the time equation:

$\displaystyle i\hbar\frac{\partial\Xi}{\partial t}=E\Xi \ \ \ \ \ (8)$

where

$\displaystyle \Xi\left(t\right)=e^{-iat} \ \ \ \ \ (9)$

Therefore

$\displaystyle E=\hbar a \ \ \ \ \ (10)$

From 7 we have

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}$ $\displaystyle =$ $\displaystyle \left(\frac{2ma}{\pi\hbar}\right)^{1/4}a\left(\hbar-2amx^{2}\right)e^{-amx^{2}/\hbar}\ \ \ \ \ (11)$ $\displaystyle V\left(x\right)$ $\displaystyle =$ $\displaystyle \frac{E\psi(x)+\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}}{\psi\left(x\right)}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2ma^{2}x^{2} \ \ \ \ \ (13)$

This is the harmonic oscillator potential, and the wave function is actually the ground state of that potential.

We can work out a few average values:

$\displaystyle \left\langle x\right\rangle =0 \ \ \ \ \ (14)$

since ${\psi\left(x\right)}$ is even.

 $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{2}\psi^{2}dx=\frac{\hbar}{4am}\ \ \ \ \ (15)$ $\displaystyle \left\langle p\right\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-\infty}^{\infty}\psi\frac{\partial\psi}{\partial x}dx=0\ \ \ \ \ (16)$ $\displaystyle \left\langle p^{2}\right\rangle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\int_{-\infty}^{\infty}\psi\frac{\partial^{2}\psi}{\partial x^{2}}dx=\hbar ma \ \ \ \ \ (17)$

The standard deviations are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}=\frac{1}{2}\sqrt{\frac{\hbar}{ma}}\ \ \ \ \ (18)$ $\displaystyle \sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}=\sqrt{\hbar ma} \ \ \ \ \ (19)$

and the uncertainty principle is

$\displaystyle \sigma_{x}\sigma_{p}=\frac{\hbar}{2} \ \ \ \ \ (20)$

so in this case, the uncertainty is the minimum possible.

## 25 thoughts on “Harmonic oscillator: statistics”

1. gwrowe Post author

Equation 11 is the second derivative of equation 6, eqn 12 is eqn 7 solved for ${V\left(x\right)}$. You can then substitute eqns 6, 10 and 11 into 12 to get 13.

1. gwrowe Post author

Actually I used the Maple software package, but if you wanted to do it by hand then you’d use integration by parts (never heard it called parametric derivatives, though, but I assume that’s what you mean).

1. Cindy

equation 15~17 using the notation {\psi\left(x\right)} (small Psi)
did you include A in it??
or i will get the same result without considering A

i did these calculation by hand using Big Psi and i don’t feel i will get the same result if i miss A @@

thank u >W<

1. gwrowe Post author

${\psi}$ is given by equation 6 which includes ${A}$.

2. Mateus

Hello, isn’t equation (15) wrong??
Shouldn’t it be: [; = \frac{ \hbar^2 \pi}{8 m^2 a^2} ;] ??
Because [; \psi ^2 = Sqrt[\frac{\hbar \pi}{2 m a}] ;]
and [; \int x^2 \psi^2 dx = [ \frac{\hbar}{2 m a} ]^(3/2) * \frac{\pi}{2} ;]

1. Mateus

Sorry for the formatation, here it goes correct:
Hello, isn’t equation (15) wrong??
Shouldn’t it be: [; = \frac{ \hbar^2 \pi}{8 m^2 a^2} ??

Because [; \psi ^2 = \sqrt{\frac{\hbar \pi}{2 m a}} exp[- \frac{2 a m}{\hbar} x^2} ) ;]

and [; \int x^2 exp[- \frac{2 a m}{\hbar} x^2} ) dx = \left[ \frac{\hbar}{2 m a} \right] ^{3/2} * \frac{\sqrt{\pi}}{2} ;]

1. gwrowe Post author

Sorry, I can’t read your Latex code properly so I’m not entirely sure what it says. However, the answer you give (${\frac{\hbar^{2}\pi}{8m^{2}a^{2}}}$) cannot be correct as you can see if you check the units. ${\left\langle x^{2}\right\rangle }$ must have the units of ${\mbox{length}^{2}}$ but your answer has units of ${\mbox{length}^{4}}$. I did the integral in equation 15 using Maple, and the units check out, so I’m pretty sure it’s right.

1. Mateus

Sorry for the format. You’re right, I was making a mistake with the constants. Sorry, and thank you, 🙂

3. Chern Xi

Dear prof, what is the reasoning behind”the wave function is actually the ground state of that potential.” ? Should I understand it as a consequence of E=0 in equation (7)?

Also, can we further deduce that since E=0, the particle is in its ground state where its uncertainty of both momentum and position are minimum, which in turns proving that (sigma x)(sigma p) must be greater than
( h-bar /2)?

1. gwrowe Post author

${E}$ is not zero (see equation 10). The statement after equation 13 just points out that the given wave function is the ground state of the harmonic oscillator, which we worked out earlier (follow the link in the post).

1. gwrowe Post author

Just plug in ${\psi}$ from equation 6. By using integration by parts, you can reduce them to standard Gaussian integrals, which you can look up in a table.

4. Joey Contreras

I think I’m missing something in regards to equation (13). When I solved equation (12) I had what you had for (13) except there was also an extra term E. So I had V(x) = 2m(a^2)(x^2) + (h-bar)(a). Did I make some mistake in my math?

5. serm

Shouldn’t the exp. val. of (equation 17) be multiplied by the normalization constant also?
(thus, = m*h_bar*a*sqrt(2*a*m/pi*h_bar))

1. gwrowe Post author

Equation 17 is correct. Just plug in ${\psi}$ from equation 6 and most of the constants cancel out.