# Heisenberg uncertainty principle

1. Dec 3, 2006

### ovoleg

Hey guys I have a few questions...

My book defines the Heisenberg uncertainty principle by

∆x∆Px >= aitch-bar

All other resources I have have it stated as
∆x∆Px >= aitch-bar/2. They mention that ∆Px and ∆x represent the rms values of independent measurements.

My book represents ∆x and ∆Px as the standard-deviation uncertanties right..

So say for instance you get a general question like x-cordinate of a proton is measured with uncertainty of 1.3mm. What is the xcomponent of velocity to the minimum percentage of uncertainty of 33%.

would you take ∆x as the standard deviation uncertanties or rms values ?

It seems like it varies from book to book but in general shouldnt this be the same? Say someone posed a question like this online, how would I know what to use?

2. Dec 4, 2006

### Jheriko

When you have:

$$\Delta x \Delta p = \hbar$$

The $\Delta$'s do not mean the standard deviation, but instead the deviation which covers 50% of the possible values. When we have:

$$\Delta x \Delta p = \frac{\hbar}{2}$$

It requires that $\Delta$ be the standard deviation.

AFAIK, both of these are only actually correct when x and p are normally distributed.

Hope this clears that up for you.

Last edited: Dec 4, 2006
3. Dec 4, 2006

### ovoleg

I'm still a bit confused as to how both of them could be correct?

Makes sense that if delta(x) is one standard deviation then in the first one it covers half the values, but then how are these two equivalent?

Also, rms does not equal a standard deviation as far as I know...

I'm confused :/

Last edited: Dec 4, 2006
4. Dec 4, 2006

### Jheriko

First off I made some nasty typos in my tex and the equalses should be greater than or equal tos.

$$\Delta x \Delta p & \ge \hbar$$

$$\Delta x \Delta p & \ge \frac{\hbar}{2}$$

Secondly, the important point is that in the first equation the delta does NOT represent a standard deviation. A standard deviation covers some percentage of the values that I can never remember... ~68% apparently. The required deviation for the first form uses the size of the range which covers
50% of the values. Its just confusing that the same symbol gets used, it might be better to have

$$\Delta_{50\%} x \Delta_{50\%} p & \ge \hbar$$

$$\Delta x \Delta p & \ge \frac{\hbar}{2}$$

This is not standard though, and just something I made up on the spot.

I don't know where the root mean square is coming from...

Last edited: Dec 4, 2006
5. Dec 4, 2006

### ovoleg

if it was 50% of each then it would be aitch-bar/4?

:( I'm lost

6. Dec 4, 2006

### Staff: Mentor

We often call it that because we define the uncertainty in x as

$$\Delta x \equiv \sqrt{<(x - <x>)^2>}$$

i.e. as the square root of the mean of the square of the difference between x and its mean. In QM of course we speak of "expectation value" rather than "mean" in this context.

I think we define $\Delta x$ this way because it has nice mathematical properties as compared to other ways that we could define it.

7. Dec 4, 2006

### ovoleg

Well as the example that I posed initially, how would you determine what the delta values are? Are they standard deviations or are they rms ?

and it doesnt seem to be as nice of a relationship when you compare rms and a standard deviation(1/2)

8. Dec 4, 2006

### Staff: Mentor

First, what are your definitions of "standard deviation" and "rms"? To me, they mean basically the same thing. But the important thing is the mathematical definition, not the words.

For the definitions of $\Delta x$ and $\Delta p$ that your book is using, you had best look for them in the book itself. Whatever they're using, it isn't the standard definition. Mathematically, the standard definition (in terms of expectation values) is the one that I gave, and it leads to $\Delta x \Delta p \ge \hbar/2$.

Either your book's $\Delta x \Delta p \ge \hbar$ is a typographical error, or the author is being a bit sloppy by neglecting numerical constants, or he's using a non-standard definition of $\Delta x$ and $\Delta p$ that is different from the standard one by a factor of $\sqrt{2}$.

9. Dec 5, 2006

### ovoleg

Thanks, actually I had to look up rms and "standard deviation" and they do differ...but not nicely for me to rearrange the equation to remove the constant.

It does seem like he has a different defition for Delta(x) and Delta(p) but the problems are worded in a way that can be interpreted with either inequality which is misleading because Im not sure which one to use...

I was just curious on who else would see this type of definition.

Thanks again!