Why can't the Uncertainty Principle be broken for individual measurements if it is a statistical law?

Question

The Heisenberg Uncertainty Principle is derived for two operators $\hat A$ and $\hat B$ as $$\Delta \hat A\ \Delta \hat B \geq \dfrac{1}{2}|\langle[\hat A, \hat B] \rangle|$$ where $\Delta$ denotes the standard deviation in a variable.

Taking position and momentum as $\hat A$ and $\hat B$ respectively, we get- $$\Delta \hat x\ \Delta \hat p_x\geq \hbar/2$$

Now, mathematically what this means is that if we prepare a large number of states $\ | \Psi \rangle$ and perform measurements of the position and momentum on them one by one, the RMS value of deviation from the mean for both $x$ and $p_x$ will show an inverse relationship with each other.

How then does this lead to a restriction on the individually measured values of position and momentum? How do we make claims such as the particle is restricted to this box so it can't have a zero momentum and so forth?

Answer 1

While uncertainties can be interpreted as statements about ensembles of identically-prepared states, that's not how they are defined. You take a state $\psi$ and then the uncertainty $\sigma_\psi(A) = \sqrt{\langle A^2\rangle - \langle A\rangle^2}$ for any observable $A$ is simply a property of that state.

A state that has a definite momentum of zero would have to be an eigenstate of momentum. You can straightforwardly show that $\sigma_\psi(A) = 0$ for an eigenstate of $A$. So any state that has a well-defined $\sigma_\psi(x)$ cannot be an eigenstate of momentum, since the uncertainty principle implies $\sigma_\psi(p)\neq 0$. Note that this argument shows that in general you cannot have eigen"states" of position or momentum if both $\sigma_\psi(x)$ and $\sigma_\psi(p)$ exist.

"It can't have zero momentum" is not supposed to mean it's impossible to measure 0 momentum - just that it's not an eigenstate where the only possible result for momentum is zero (or indeed any other value of momentum).

In somewhat sloppy phrasing, "It can't have zero momentum" might also refer to the expectation value of the magnitude of momentum: If $\sigma_\psi(p)\neq 0$, then since $\sigma_\psi(p)\leq \sqrt{\langle p^2\rangle}$ we have that $\langle p^2\rangle$ is also non-zero.

Answer 2

(This first section refers to v1 of the post)

Now, mathematically what this means is that if we prepare a large number of states $|Ψ⟩$ and perform measurements of the position and momentum on them (not doing >1 measurement on any prepared system)...

First mistake. For example, measuring $X$ on one state gives you a new state, so the resulting $\Delta p$ you get from subsequent momentum measurements will not be related (by the HUP) to the previous $\Delta x$ you obtained.

The HUP is a statement about measurements of $X$ or $P$ separately on a large number of similar states. So, take a large number of similar states and measure $X$ for each one to get $\Delta x$. Take another larger number of similar states and measure their momentum to get $\Delta p$. The HUP says that no matter what state you started with, you will always find $\Delta x\cdot\Delta p$ to be no smaller than $\hbar/2$.

...the RMS value of deviation from the mean for both will show an inverse relationship with each other.

Second mistake. If you have multiple similar states and do the above steps you will get a single $\Delta x$ and a single $\Delta p$. There is no inverse relationship to be found.

However, let's say we were to then take a different state and do the same procedure, and let's say $\Delta x$ is now smaller than before. We still cannot say anything about the new $\Delta p$. $\Delta p$ could increase or decrease; all we would know is that $\Delta x\cdot\Delta p$ can be no smaller than $\hbar/2$. However, if the previous $\Delta x\cdot\Delta p$ was exactly equal to $\hbar/2$, then we could guarantee a larger $\Delta p$ because the HUP must hold.

How then does this lead to a restriction on the individually measured values of position and momentum?

It doesn't. The HUP does not say anything about individual measurements.

How do we make claims such as the particle is restricted to this box so it can't have a zero momentum and so forth?

You can certainly make a $0$ momentum measurement. That doesn't violate the HUP. The HUP would be violated if $\Delta p=0$, i.e., if every momentum measurement of our large collection of similar states gave us the same value every single time.

To address the title then:

Why can't the Uncertainty Principle be broken for individual measurements if it is a statistical law?

As stated above, the HUP does not say anything about individual measurements. It just puts a bound on the product of the "spreads" of two types of measurements.

Answer 3

All measurements in QM are implied to be the ensemble measurements, i.e., they are done on many particles/systems "prepared" in the same way. A measurement perturbs the system, so it can never be done on the same system twice. Measuring two non-commuting operators on the same system would require two measurements, so it cannot be done on one system. But consecutive measurements may indeed violate the uncertainty principle, before we collect enough data for reliable statistical inference.

Update
It is necessary to distinguish between the mathematical and the sample means. Thus, mathematically we define the mean and the variance of a quantity as $$ \langle A\rangle = \int dx \psi^*(x)\hat{A}\psi(x),\\ \langle (\Delta A)^2\rangle = \int dx \psi^*(x)(\hat{A}-\langle A\rangle)^2\psi(x) $$ The uncertainty relation applied to such mathematical quantities is a rigorous mathematical statement, allowing no ambiguity.

In experiment however we will have measurements $\mathbf{A}=(A_1, A_2, ..., A_N)$ and calculate sample averages $$ \overline{x} = \frac{1}{N}\sum_{i=1}^Nx_i,\\ var(x) = \overline{(x -\overline{x})^2} = \frac{1}{N-1}\sum_{i=1}^N(x_i-\overline{x})^2 $$ These averages are themselves random quantities, which approach the mathematical averages only in the limit $N\rightarrow\infty$ (Note factor $1/(N-1)$ in the definition of the invariance - if it were $1/N$, the estimate would be biased, i.e., it would never approach the theoretical value). Thus, for finite $N$ it is quite possible to have a situation where the sample variances do not satisfy the uncertainty relation.

To summarize: the physical meaning of the uncertainty relation is statistical. More specifically:

• it is valid only for an ensemble of measurements
• it cannot be used for a single measurement (moreover, two non-commuting quantities cannot be measured simultaneously)

Answer 4

Quantum mechanics cannot be reduced to the HUP. There are some basic axioms of QM and the HUP follows from them. The relevant axioms here are:

1. Quantum states are vectors $\psi$ in a Hilbert space $\mathbf{H}$.
2. Observables are operators $A: \mathbf{H} \rightarrow\mathbf{H}$
3. The measurement outcomes can only be the eigenvalues of the corresponding operator.
4. The measurement outcome is random (that means unknowable in general). The probability of measuring the value $a$ is given by the absolute square of the overlab $\langle \phi_a |\psi \rangle$ of the quantum state and the eigenvector of $A$, corresponding to the eigenvalue $a$

Therefore the value of an observable for a given state is only known, if the quantum state is an eigenstate of that observable (for example immediately after a measurement). It follows that we can know the value of two observables simultaneously only if the state is an eigenstate of both observables at once. This is only possible if the two observables commute.

Answer 5

How do we make claims such as the particle is restricted to this box so it can't have a zero momentum and so forth?

Have you seen this claim made? It is at best shorthand for something else, and at worst just wrong. First, one should talk of the expected value or measured value of a particle, rather than speaking of it just having a value. The expected value of the momentum of a particle confined to a box absolutely can be zero; if it's confined to a box, and the box isn't moving, then overall expected value has to be zero (there are states where the expected value varies over time, but the average is still zero).

The probability of the measured value of its momentum being zero is zero (that is, $p(\text{momentum}=0)=0$), but that's just because the probability of it being any particular value is zero. The probability density for its momentum will, for normal states, be highest at zero.

Another correct statement would be that the momentum can't be restricted to zero.