# Gaussian function to derive Heisenberg's uncertainty principle

1. Feb 24, 2013

### 71GA

At our QM intro our professor said that we derive uncertainty principle using the integral of plane waves $\psi = \psi_0(k) e^{i(kx - \omega t)}$ over wave numbers $k$. We do it at $t=0$ hence $\psi = \psi_0(k) e^{ikx}$

$\psi = \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k$

where $\psi_0(k)$ is a $k$-dependent normalisation factor (please correct me if I am wrong). This dependency was said to be a Gaussian function

$\psi_0(k)= \psi_0 e^{i(k-k_0)^2/4\sigma k^2}$

where $\psi_0$ is an ordinary normalisation factor (please correct me if I am wrong).

QUESTION 1: Why do we choose $\psi_0(k)$ as a gauss function? Why is this function so appropriate in this case?

QUESTION 2: I don't know how did our professor get a gauss function with an imagnary number $i$ in it. His gauss is nothing like the one on Wikipedia which is

$f(x) = a e^{-(x-b)^2/2c^2}$

QUESTION 3: We used the first integral i wrote down to calculate the Heisenberg's uncertainty principle like shown below, but it seems to me that most of the steps are missing and this is the reason i don't understand this. Could anyone explain to me step by step how to do this.

$\begin{split} \psi &= \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k\\ \psi &= \int\limits_{-\infty}^{+\infty} \psi_0 e^{i(k-k_0)^2/4\sigma k^2} \cdot e^{ikx} \, \textrm{d} k\\ \psi &= \psi_0 2 \sqrt{\pi} e^{ik_0x} e^{-x/2 \sigma k^2} \end{split}$

I think this is connected to a Gaussian integral, but it doesn't look quite like it to me. Well in the end our professor just says that out of the above it follows that

$\boxed{\delta x \delta k = \frac{1}{2}}$

I don't understand this neither. It was way too fast or me.

2. Feb 24, 2013

### Staff: Mentor

1 and 2: This choice gives something which looks like a moving particle. Without a careful choice of $\psi(k)$, you get some wave spread out over the whole space.

It is one. You can combine those two exponentials to a single one, make a linear substitution and get the gaussian integral. Are you sure that the last exponent in the last line has just an x there?

For δxδk >> 1/2, the last exponential function should vanish, so you get a significant wave only if the product δxδk is small. A better analysis would give that factor of 1/2.

3. Feb 24, 2013

### andrien

uncertainty product is minimum for gaussian wave packet.the inequality becomes the equality.

4. Feb 24, 2013

### 71GA

So what does this mean our professor was wrong? Could please someone show a correct derivation f possible using equations.

Do you mean this one?

$\begin{split} \psi &= \psi_0 2 \sqrt{\pi} e^{ik_0x} e^{-x/2 \sigma k^2} \end{split}$

Thats how our professor did this. Was he wrong? I am totaly confused.

Last edited: Feb 24, 2013
5. Feb 24, 2013

### Staff: Mentor

I'm not sure, but the equation looks not as I would expect it.

6. Feb 24, 2013

### 71GA

I think my professor was wrong so please tell me how would you expect the equation to be.

7. Feb 24, 2013

### Staff: Mentor

It is better if you look for a quantum mechanics book with a proper derivation of the uncertainty for a gaussian wave packet.

8. Feb 24, 2013

### 71GA

Well i am quite weak in mathematics. Is there any nice book suited for me?

9. Feb 24, 2013

### tom.stoer

The HUP is a property of the Fourier Transformation and applies to every pair f(x), g(p) where g is the FT of f.

Four Gausssian wave pakets the product Δx Δp is minimized.

But of course one cannot derive the HUP in general based on a special choice of functions.

10. Feb 24, 2013

### 71GA

Could you provide such a case and derivaton for HUP?

11. Feb 24, 2013

### tom.stoer

Try this derivation which does not use Fourier transform at all but only "algebraic" formulas applied to Hilbert space vectors

12. Feb 25, 2013

### andrien

In the case of a special wave function as the given gaussian one,one can do get a relation by exploiting the definition of Δx and Δp as
Δx=√(<x2>-<x>2) and similarly for Δp.After that one can employ
<x2>=∫ψ*x2ψ dx(in case of noramalized ψ)
one can evaluate <x2>,<x>,<p>,<p2> by this using first quantized form for momentum operator i.e. while evaluating integral of p2,put -d2/dx2.then one can evaluate Δx and Δp and product of it will give h-/2
p=h-k,gives ΔxΔp=1/2.

13. Feb 25, 2013

### stevendaryl

Staff Emeritus

$(\delta x)^2$ = $\langle(x - \langle x \rangle)^2\rangle$
$(\delta p)^2$ = $\langle(p - \langle p \rangle)^2\rangle$

where $\langle A \rangle = \int \psi^{*}(x) A \psi(x) dx$

Then we can prove (using the calculus of variations) that the $\psi(x)$ that minimizes $\delta x \delta p$ is a Gaussian.

14. Feb 25, 2013

### Staff: Mentor

I think the second equation should be

$$\psi_0(k) = \psi_0 e^{-(k-k_0)^2/4\sigma^2}$$

with a "-" sign instead of an "i" in the exponent, and no "k" in the denominator. That "k" could be a subscript:

$$\psi_0(k) = \psi_0 e^{-(k-k_0)^2/4\sigma_k^2}$$

which would be appropriate because $\sigma$ or $\sigma_k$ is the standard deviation of k.

15. Feb 25, 2013

### 71GA

I think that our professor was lecturing by using this document where they used a suare root of a gauss so this is where he got $\psi_0(k)$:

$\psi_0(k)= \sqrt{gauss} = \sqrt{\psi_0 e^{-(k-k_0)^2/2\sigma_k^2}} = \psi_0 e^{-(k-k_0)^2/4\sigma_k^2}$

My question is Why do we take a square root of a gauss?

Last edited: Feb 25, 2013
16. Feb 25, 2013

### 71GA

Our professor did a mistake. You are right! Please take a look at my previous post.

17. Feb 25, 2013

### Staff: Mentor

As you probably know already, $\psi(x)$ is the probability amplitude for x, which we "complex-square" to get the probability distribution for x: $P(x) = |\psi(x)|^2$.

Similarly, your $\psi_0(k)$ is the probability amplitude for k, which we "complex-square" to get the probability distribution for k: $P_k(k) = |\psi_0(k)|^2$.

I suspect that your professor wants to make $P_k(k)$ a Gaussian with standard deviation $\sigma_k$. Some books do it the other way, i.e. they make $\psi_0(k)$ a Gaussian with standard deviation $\sigma_k$.

Last edited: Feb 25, 2013
18. Feb 25, 2013

### 71GA

In the link i provided earlier for example they do both ($\psi_0(k)$ as well as $\psi_0(x)$) a standard deviation - Gauss. Why is that so?

Last edited: Feb 25, 2013
19. Feb 26, 2013

### andrien

it is much easier than this.The schwarz inequality becomes an equality when the two function f and g used are in relation f=λg,where λ is a constant.Also f and g are most appropriately chosen as for deriving uncertainty principle
f=-ih-∂ψ/∂x,g=ixψ,the condition f=λg does give a gaussian function.

20. Feb 26, 2013

### 71GA

But i am weak in math and Fourier is more suiable for me. I hae no clue about Schwarz inequaity... Could anyone provide a clean derivation of uncertainty principle using gauss wave packet together with Fourier?