# Prove Heisenberg Uncertainty Principle for Ground State Harmonic Oscillator

1. Oct 6, 2012

### Sciencenerd3

Ground State Wave Equation:
ψ0=(a/∏)(1/4)e(-ax2/2)

Prove the Heisenberg Uncertainty principle ≥h(bar)/2 by way of expectation values.

First I found <x>=0 because it was an odd function

then I found <Px>=0 because it was an odd function

Then <x2>=∫(a/∏)(1/2)x2e(-ax2)/2dx=1/2a by way of even function integral table

The problem I am having difficulty on is finding <Px2>
It involves taking the double derivative and then integrating using a table (for me). I need to go through all of the derivatives and integrals I believe in order to get things to cancel so that only h/2 will be left over but I keep getting 0 for <Px2>

I have literally been trying to work this out for ≈4-5 hrs so any help anyone can offer would be extremely appreciated.

2. Oct 6, 2012

### The_Duck

Here's a shortcut: H = p^2/2m + 1/2 m w^2 x^2, so <H> = (1/2m)<p^2> + (mw^2/2)<x^2>. You know <H> and <x^2>, so you can solve for <p^2>.

3. Oct 6, 2012

### Sciencenerd3

This seems as though it would work but I have to solve it in a specific way. My answer would be close if too many things were not cancelling but I cannot seem to work it out fully. After I integrate after taking the double derivative and keeping in mind to used the product rule, I am getting zero.

4. Oct 6, 2012

### vanhees71

As far as I understand you simply are supposed to calculate $\langle x \rangle$, $\langle p \rangle$, $\langle x^2 \rangle$, $\langle p^2 \rangle$. Since you've given the wavefunction in position space, you simply have to use the operators
$$\hat{x}=x, \quad \hat{p}=-\mathrm{i} \hbar \partial_x$$
and use the general equation for expectation values
$$\langle A \rangle=\int_{\mathbb{R}} \mathrm{d} x \; \psi^*(x) \hat{A} \psi(x).$$
Of course you must also check that the wave function is normalized properly, i.e., that
$$\int_{\mathbb{R}} \mathrm{d} x \; |\psi(x)|^2=1.$$

5. Oct 7, 2012

### Sciencenerd3

Yes, the wavefunction is normalized because it is the ground state harmonic oscillator wavefunction. However, when I try to use the expectation value equation:

⟨A⟩=∫ℝdxψ∗(x)Aˆψ(x).

I get <P2>=h2(bar)/i2[-2a+2a]=0. However, I know that this value is not 0.

I am supposed to use the expectation equation and values to get Δx and Δp and show ΔxΔp≥h(bar)/2.

Uncertainty: Δx=√(<x2>-<x>2) but <x>=0. This is a similar result for Δp because <p>=0 as well.

Then after this you should be able to multiply both uncertainty values to get the Heisenberg uncertainty. But not matter what I do I can not get the expectation value of <p2> not to equal zero. And I also get <x2>=1/2a.

Is there anyone who can run through the expectation of momentum squared. It involves taking the double derivative and then integrating over allspace for the given wavefunction:

ψ0=(a/∏)(1/4)e(-ax2/2)

6. Oct 7, 2012

7. Oct 8, 2012

### Sciencenerd3

$\psi_0=(\frac{\alpha}{\pi})^\frac{1}{4}e^{-\frac{\alpha x^2}{2}}$

So using the expectation equation:

$<p^2>=∫\psi_0^*\hat{p}\hat{p}\psi_0dx$

$<p^2>=∫\psi_0^*\frac{\hbar^2}{i^2}\frac{∂^2}{∂x^2}\psi_0dx$

$<p^2>=∫\frac{\hbar^2}{i^2}\frac{∂^2}{∂x^2}(\frac{ \alpha}{\pi})^\frac{1}{2}e^{-\alpha x^2}$

$\frac{∂^2}{∂x^2}(\frac{ \alpha}{\pi})^\frac{1}{2}e^{-\alpha x^2}=\frac{∂}{∂x}((\frac{ \alpha}{\pi})^\frac{1}{2}\cdot-2\alpha xe^{-\alpha x^2})=(\frac{ \alpha}{\pi})^\frac{1}{2}\cdot-2\alpha e^{-\alpha x^2}+(\frac{ \alpha}{\pi})^\frac{1}{2}\cdot4\alpha^2x^2 e^{-\alpha x^2}$

$\int_{-∞}^∞e^{-\alpha x^2}=(\frac{\pi}{\alpha})^{\frac{1}{2}}$

$2\cdot\int_0^∞x^2e^{-\alpha x^2}=(\frac{\pi}{\alpha})^{\frac{1}{2}}\cdot\frac{2}{4\alpha}$

So putting these values in for the integrals, I get:

$<p^2>=\int_{-∞}^∞[(\frac{ \alpha}{\pi})^\frac{1}{2}\cdot-2\alpha e^{-\alpha x^2}+(\frac{ \alpha}{\pi})^\frac{1}{2}\cdot4\alpha^2x^2 e^{-\alpha x^2}]=\frac{\hbar^2}{i^2}[-2\alpha+2\alpha]=0$

I cannot figure out if I'm doing the derivatives wrong or the integrals or neither. Maybe my values from the integral tables were figured incorrectly. Whatever it is, it has evaded my solving for a long time now.

8. Oct 8, 2012

### Sciencenerd3

Could it be that I am acting on both wave functions multiplied together with the momentum operator instead of just the second wave function in the expectation equation?

9. Oct 8, 2012

### tom.stoer

Here's the wrong step

The first equation is OK, but then you move ψ* from the left of the differential operator to the right, such that ∂x² is no longer acting on ψ but on |ψ|².

But

$$\int_{-\infty}^{+\infty}dx\, \psi^\ast \frac{\partial^2}{\partial x^2} \psi \neq \int_{-\infty}^{+\infty}dx\, \frac{\partial^2}{\partial x^2} |\psi|^2$$

This integral is zero (trvially) b/c

$$\int_{-\infty}^{+\infty}dx\, \frac{\partial^2}{\partial x^2} |\psi|^2 = \left. \frac{\partial}{\partial x} |\psi|^2\right|_{-\infty}^{+\infty} = 0$$

which vanishes due to the asymptotics of ψ.

What you can do is

$$\int_{-\infty}^{+\infty}dx\, \psi^\ast \frac{\partial^2}{\partial x^2} \psi = - \int_{-\infty}^{+\infty}dx\, \left( \frac{\partial}{\partial x} \psi^\ast\right) \left( \frac{\partial}{\partial x} \psi\right) = \int_{-\infty}^{+\infty}dx\, \left( \frac{\partial}{\partial x} \psi\right)^2$$

where I first used integration by parts (which is always OK for wave functions vanishing at infinity) and then ψ* = ψ (which holds only in special cases).

10. Oct 8, 2012

### Staff: Mentor

That's your problem... $\partial^2 / \partial x^2$ applies only to $\psi_0$, not to $\psi_0^*$:

$$\langle p^2 \rangle = \int {(\psi_0^*) \left( -\hbar^2 \frac{\partial^2 \psi_0}{\partial x^2} \right) dx}$$

(I see tom.stoer beat me to it while I was debugging my LaTeX.)

11. Oct 8, 2012

### Sciencenerd3

Thanks so much. i cannot believe I wasted so much time on something so simple and trivial like forgetting order of operations. I'm new to the forum, is there any specific way I am supposed to give thanks or credit?

12. Oct 8, 2012

### Staff: Mentor

We'll consider it given. At least I will... maybe Tom will have a special request like a bottle of beer or something.

13. Oct 8, 2012

### tom.stoer

in such cases a bottle of whisky would be appropriate