# Different forms of Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle is often written in two forms:

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

and

$$\sigma_x \sigma_p \geq \frac{\hbar}{2}.$$

Are these two equivalent? I've been told they are, but it doesn't make sense to me. For instance, in an infinite square well, $$\Delta x =$$ the width of the well. Whereas $$\sigma_x$$ is equal to the standard deviation of the wave function.

If they are equivalent, then $$\Delta x = \sigma_x = \sqrt{\langle x^2\rangle - \langle x\rangle^2}$$ meaning we can find the value of $$\langle x^2\rangle$$ using:

$$\langle x^2\rangle = (\Delta x)^2 + \langle x\rangle^2$$

Does this make sense? Or have I been misled about the equivalence of the two forms of the HUP?

• More on HUP & statistics Aug 10, 2020 at 10:19
• Why do you think that $\Delta x$ does not denote the standard deviation as well? Aug 10, 2020 at 10:20
• @ACuriousMind Until just now, I thought $\Delta x$ was equal to the full width of the well, which I've learned below is not the case. It makes sense now too as it's an uncertainty in x. So I had thought if it's equal to the full width of the well, how could it be equal to the standard deviation when the SD is not the full width of the well. Aug 10, 2020 at 11:05

The statistical interpretation of quantum mechanics tells us that the "best" that we can know a priori (i.e. before carrying out a measurement, an experiment) from a theoretical study of a physical system is, in general, a range of possible values. Since you have a range of possibilities, the way naturally opens up for a statistical analysis: you have a distribution of values ​​characterized by an average value and a dispersion, $$\sigma$$, around it.The product of the two $$\sigma$$ associated with the distribuitions of two coniugate osservables can't go below the value indicated in the HUP.

If we instead carry out an experiment and successive measurements of two conjugate quantities, "each time returning the system to the $$\Psi$$ preceding the measurements", A and B, we obtain different values ​​characterized by uncertainties $$\Delta A$$ and $$\Delta B$$ whose product will have an upper limit. As De Broglie said, we are therefore dealing with pre-measurement (in the first case) and post-measurement (in the second) uncertainty relations.

For instance, the infinite square well centered in the origin, the particle can occupy all positions between -L/2 and +L/2: so the average value is x=0 and the dispersion is L/2. Or, if you performe a large number of measurements, you'll obtain the uncertainty, $$\Delta x$$, is L/2 for the mean value x=0.

• Ah thankyou @Francesco, I was taking $\Delta x$ to be equal to the entire width of the well, so L, rather than L/2! Aug 10, 2020 at 11:04
• @Franceso, does the last part of my question make sense to you? Provided I take $\Delta x$ to be L/2, I can use this to find $<x^2>$? (Given I already know the value of $<x>^2$) Aug 10, 2020 at 11:09
• @jstook you're welcome. Sorry, I forgot to tell you that I used a gross value for $\Delta x$ just for fix ideas. The $\sigma_x$ has a different and smaller value. Anyway, if you know the exact value of $\Delta x$ you can use the definition of standard deviation to calculate the mean value of $x^2$ noting that it's the mean value you obtain if you perform a big number of measurement on our system. Aug 10, 2020 at 14:15
• Thanks again @Francesco. I think I'm back at my original confusion though. My lecturer has told me the two forms, using sigma and delta notation are the same and can be used interchangeably. However my confusion stemmed from the fact that I thought $\sigma_x$ will be smaller than $\Delta x$ which you have mentioned above. If this is the case, how can we use the two interchangeably? And for an infinite square well, are you saying $\Delta x$ IS equal to the L/2 mentioned above? Or by using the wording gross value, you're just conveying the idea? In my case, my well is centered about 0. Aug 10, 2020 at 21:10
• Which is easy enough to do of course, but if I knew $\sigma_x$ I could then re arrange to solve for $<x^2>$ and use it to find $<p^2>$ as well. I'm now, after reading your responses and others on this page thinking that we can't just give a value to $\sigma_x$ without finding all the expectation values? Or to be clearer, there isn't an 'accepted' value within the infinite square well that can just be stated without proof? Aug 11, 2020 at 23:00

Most general form of uncertainty principle is : $$\boxed{\sigma _{A}\sigma _{B}\geq \left|{\frac {1}{2i}}\langle [{A},{B}]\rangle \right|}$$

Where $$A,B$$ is Hermitian operators which do not commute, namely their commutator : $$[A,B]={A}{B}-{B}{A} \neq 0$$

If that's the case, then it's impossible to measure variables $$A,B$$ simultaneously.

Now let's try to find out commutator $$[x,p]$$, for that we need to solve expression $$xp~\psi(x) -px~\psi(x)$$

Substituting quantum mechanical momentum operator and re-arranging terms gives : $$i\hbar\frac{\partial}{\partial x} \left(x\psi(x)\right) - xi\hbar\frac{\partial}{\partial x}\psi(x)$$

Using multiplication rule for first term gives : $$i\hbar\psi(x) + xi\hbar\frac{\partial}{\partial x}\psi(x) - xi\hbar\frac{\partial}{\partial x}\psi(x)$$

Noticing that the last two terms cancels each other, so we get final relationship:

$$[x,p]\psi(x) = i\hbar~\psi(x)$$

This means that our position-momentum commutator is :

$$[x,p]=i\hbar$$

Substituting resulting commutator back into general uncertainty principle form we get: $$\sigma _{x}\sigma _{p}\geq \frac{\hbar}{2}$$

• Cheers Agnius, I hadn't seen it like that. My QM course has skipped a lot of this sort of background! Aug 11, 2020 at 22:47