# Heisenberg uncertainty principle and zero point energy

On my book is written: for a particle in an infinite square well (supposed 1D and large $a$) \begin{align} \Delta x\sim a \to& \text{Heisenberg uncertainty principle} \\ \to& \Delta p _\text{min}\sim\frac{h}{2\pi a} \\ \left( E=\frac{p^2}{2m} \right) \to& E_\text{min}\sim\frac{h^2}{8\pi^2a^2m}\,. \end{align} However, I'm not sure that the last passage is legal: how is it possible to consider $\Delta p _\text{min}$ and $p$ the same thing? The first is the standard deviation of the aleatory variable $P$, while the second one is the physical value of momentum, so I'd like to understand if there is another way to determine the zero point energy of a system only using the Heisenberg uncertainty principle.

• Possible duplicate of Heisenberg's uncertainty principle - $\Delta p$ – AccidentalFourierTransform Aug 25 '17 at 22:39
• Δp is the smallest measurable increment of p. The lowest mathematical value of p is zero, for p is p=0, thus p+Δp =Δp. In physics notation one uses the delta to differentiate between "an interval" and "a standard deviation". for the standard deviation we use dp – anna v Aug 26 '17 at 4:03
• @annav I don't think that's true. Formally, the $\Delta p$ in the Heisenberg relationship is exactly the standard deviation. When you derive Heisenberg uncertainty, what pops out are the standard deviations of $x$ and $p$, not their "smallest measurable increments." – Jahan Claes Aug 29 '17 at 19:46
• @JahanClaes sorry you are wrong, unless you have an expanded defintion of standard deviation. If it were the standard deviation it would imply all quantum phenomena were random following the poisson or gaussian. The probability distribution for quantum phenomena are squares of solutions of boundary value problems for the QM differential equations, not gaussian. Thus in physics we use this notation to separate gaussians from the QM distributions. In effect, the smallest measurable increment gives a handle on the QM distribution function. – anna v Aug 30 '17 at 3:51
• @annav I'm sorry, but I'm correct. See, e.g., Shankar Chapter 9. He proves that for any state $|\psi\rangle$, you can define $\Delta p=\langle \psi| p^2|\psi\rangle-\langle \psi|p|\psi\rangle^2$ and $\Delta x$ similarly, and that you may then prove $\Delta p\Delta x\geq \hbar/2$. It is possible to define a standard deviation in QM, as I just showed, and it turns out it is useful. Nothing about using a standard deviation implies a distribution is necessaraily Gaussian; every probability distribution has a standard deviation. – Jahan Claes Aug 30 '17 at 4:32

One way to think of this is in terms of expectation values. When you say $\Delta p$, what you really mean is the standard deviation of $p$.

$$\Delta p = \sqrt{\langle p^2\rangle-\langle p \rangle^2}$$ In the case of the ground state, you expect $\langle p \rangle=0$ by symmetry, so you just have $\Delta p = \sqrt{\langle p^2\rangle}$. Then you can consider the expectation value of the energy,

$$\langle E\rangle = \langle \frac{p^2}{2m}\rangle=\frac{\langle p^2\rangle}{2m}=\frac{(\Delta p)^2}{2m}$$

So far, everything we've written has been exact. But we want to find the minimum possible value for the energy. A moments thought should tell you that the minimum of $\langle E\rangle$ and the minimum of the energy coincide. So you try to find the smallest possible $\langle E\rangle$, and call that $E_{\min}$. That means you want to find the smallest possible $\Delta p$. But of course you know $\Delta x \lesssim a$, so the smallest $\Delta p$ is $\tilde{}\frac{h}{2\pi a}$. Plugging that in gives you $E_\min$.

They key is realizing that if $\langle p\rangle=0$, then the expectation value of $p^2$ is exactly $(\Delta p)^2$. Of course, everything after that is just approximations, but sometimes they work pretty well!

• Ok, but I don't understand why I should put $E[p]=0$ – Landau Aug 29 '17 at 17:17
• @Landau There's a few different ways to justify that. One is simply by symmetry. Since the square well is symmetric, the ground state should have $\langle p\rangle=0$, since any other value of $p$ would violate symmetry. After all, if $\langle p\rangle$ is positive in the ground state, that means there's something special about the positive direction, which isn't true: the square well looks the same in the positive and negative directions. – Jahan Claes Aug 29 '17 at 19:38
• @Landau Another way to say it is that you know the particle in the ground state isn't going anywhere: the expectation value of $\langle x \rangle$ isn't changing, otherwise we wouldn't be in the ground state. There's a formal way to relate $\langle p\rangle$ and $\langle x \rangle$, and it gives exactly what you'd expect: $d\langle x\rangle/dt = \langle p\rangle/m$. So if you want to make sure $\langle x \rangle$ doesn't change, you need to have $\langle p \rangle=0$. – Jahan Claes Aug 29 '17 at 19:43
• @Landau In fact, it's a generally true statement that any localized eigenstate has $\langle p \rangle =0$. – Jahan Claes Aug 29 '17 at 21:12

It's not a rigorous derivation, it's an estimation that happens to give the right result. The basic idea is that the minimum possible uncertainty in momentum is going to be of the same order as the minimum possible value of the momentum. This isn't always true, but it's often true enough.

In fact, note that the book conveniently used $\Delta x \Delta p \sim \hbar$ instead of $\Delta x \Delta p \sim \hbar/2$ to get the right result.

As for deriving the energy using only the uncertainty principle, I don't think it's possible. The HUP is just an inequality, the actual uncertainties could be larger than their minimum allowed values. Not to mention the uncertainty in some observable is not necessarily the same as its value.