Heisenberg's uncertainty isn't about measurement

Question

If we are given two operators $\hat A $ & $\hat B$ corresponding to the physical quantities $A$ & $B$, then for a given wave function $\Psi$ we know that the average values of those quantites over all possible measurements of the particle are:

$$\langle A \rangle=\langle \Psi|\hat A|\Psi \rangle=\int_V \Psi^* \hat A\Psi dV \\ \langle B \rangle=\langle \Psi|\hat B|\Psi \rangle=\int_V \Psi^* \hat B\Psi dV$$

Now if we define the operators $\hat D_A = (\hat A-\langle A\rangle \hat I)$ & $\hat D_B = (\hat B-\langle B\rangle \hat I)$ we can easily show that:

$$\langle D_A \rangle=\langle A^2 \rangle - \langle A \rangle^2=\sigma^2(A) \\ \langle D_B \rangle=\langle B^2 \rangle - \langle B \rangle^2=\sigma^2(B)$$

So The average of this newly defined variance operator indeed gives the variance of the physical quantities fot the given $\Psi$. So I imagine that it's a way to measure how "spread out" are the quantities for this exact wave function.

Now the Heisenberg uncertainty principle states that:

$$\sqrt{\langle D_A \rangle \langle D_B\rangle}=\sigma(A)\sigma(B)\geq \frac{1}{2}|\langle C\rangle |\\ \text{Where }\ \ \ i\hat C=[\hat A,\hat B]$$

Commonly in mainstream science communication we are told that the Heisenberg uncertainty principle states that (for example in position and momentum) those physical quantities can't be MEASURED both with exact precision and this is certainly a consequence of the math. However I believe the uncertainty principle is deeper than that. The way I see it is that it shows us the maximum degree of certainty we can have about both quantities before we measure the system. So for example we can have a wave function that consists of only one eigenvector of $\hat A$ so we will know the exact value of $A$ even before measuring, and this would mean that we for sure can't know anything about the value of $B$ before we measure it since it would need to have infinite standard deviation. Is my way of thinking about this correct? (This question is about the underlying concept of the Heisenberg uncertainty principle, not a homework-like question)

Answer 1

You are mostly correct, but so is the comment from Roger V. which states that the wavefunction is not itself directly physical but is a way to organize calculations about physical stuff.

I suggest that the first stage to understanding Heisenberg Uncertainty Principle (HUP) is to consider the purely classical case of a classical wave having frequency and duration. By Fourier analysis one learns that there is a lower bound on the combination $\sigma_\omega \sigma_t$ where the sigmas refer to standard deviation. In this case there is no need to introduce the concept of measurement or knowledge. In the classical case in principle everything can be measured precisely, even the very oscillations of the wave, but a wave packet of finite duration simply does not have and cannot have a perfectly well-defined frequency.

Coming now to quantum theory, there are two new considerations. First, we are dealing with wavefunctions (or state vectors etc.). Secondly, the uncertainty concerns quantities such as position and momentum, which previously were not thought to have wavelike properties, but now we know they do. By analogy with the classical case, it is correct to say that a particle simply does not have a well-defined position and momentum at the same time, no matter whether anyone measures it. In this sense the uncertainty is in the quantum system itself, not in anyone's knowledge of it. However, if one wants to say carefully what a wavefunction or a ket or a state vector is, in the sense of what role it plays in physical understanding, or how it relates to physical goings-on, then issues around knowledge are liable to come into the discussion. This is because whereas some people think a state vector simply encodes "what is so", other people think a state vector is a mathematical tool which can be used to calculate the probabilities of what may be so.

Answer 2

Commonly in mainstream science communication we are told that the Heisenberg uncertainty principle states that (for example in position and momentum) those physical quantities can't be MEASURED both with exact precision and this is certainly a consequence of the math. However I believe the uncertainty principle is deeper than that.

(emphasis is mine)
This is a misinterpretation of what the mainstream science communication says. More specifically, it says that the Heisenberg uncertainty principle states that (for example in position and momentum) those physical quantities can't be MEASURED both with exact precision. (period) This is a property of the nature, that we know from the experiments - not a consequence of the math that we use to describe the nature.

And this is the main flaw in the reasoning presented in the Q. - thinking of physical phenomena as a consequence of the math, rather than the other way around. Physical phenomena happen in certain ways, regardless of what kind of math we use to describe them, and whether we know of them or not. We could describe the measurement without resorting to concepts of wave function and standard deviation, but if this description failed to predict that position and momentum cannot be precisely measured simultaneously, the description would be wrong - because it would disagree with the experimental evidence.

The way I see it is that it shows us the maximum degree of certainty we can have about both quantities before we measure the system. So for example we can have a wave function that consists of only one eigenvector of $\hat A$ so we will know the exact value of $A$ even before measuring, and this would mean that we for sure can't know anything about the value of $B$ before we measure it since it would need to have infinite standard deviation. Is my way of thinking about this correct? (This question is about the underlying concept of the Heisenberg uncertainty principle, not a homework-like question)

Again, if we translate this from math to a physical language, we can prepare a system in a state, where a specific quantity can be measured with very high precision (we can never assure infinite precision in a finite-time experiment, and no experiment can last longer than the Universe exists.) We then can measure the conjugate quantity, and discover that we cannot predict its value. More precisely, we prepare ensemble of systems in identical states, and discover that the variation of the conjugate quantity is very large, and limited by the uncertainty principle (or something close.)

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

Answer 3

There are two problems (that are related) that I see with your proposition.

1. The information you are talking about:

maximum degree of certainty we can have about both quantities before we measure the system

is encoded in the wave-function, and if you break it down, for the observable $A$ it is in the quantity $\sigma(A)$ alone. This quantity exists and can be calculated independently of the uncertainty principle. It especially is independent of any other observable $B$ (that you could choose abitrarily).

2. The uncertainty principle yields a lower bound for $\sigma(A)$, in sofar it answers your question on the maximum degree of certainty. It is however an estimation that answers this question with an abitrarily other choosen measurement $B$, and this estimation always involves $\sigma(B)$. As such, the uncertainty principle doesn't give you any information that can be regarded a property of observable $A$ alone. Instead, it will be a property of the measurement, because it answers the question on what would happen if you want to measure $A$ and $B$ simultaneously.

Answer 4

I would interpret that mainstream science communication rather differently. I think we can all agree with how QM works, and the only point of contention is in the wording, especially the word measurement.

Let's consider the sequence of events:

1. You do position measurement and get $x$
2. Wavefunction collapses; particle is confined within $x \pm \Delta x$, the spread $\Delta x$ determined by apparatus resolution
3. Immediately follow up with a momentum measurement
4. Cannot pinpoint momentum state to better than $\Delta p \sim 1/\Delta x$ no matter what.

This is what I see the sentence

physical quantities can't be MEASURED both with exact precision

describes. No, I do not take it as the uncertainty principle, but merely a consequence of the uncertainty rinciple that is digestable by a non-physicist. And I take the position that science communications aimed at the general publics always leave many things lost in translation, often even allowing the main message to be dropped.

Answer 5

There are major problems with the interpretation of the Heisenberg relation. Actually, if there are two non-negative quantities such that x y >= 1, then neither x nor y can be zero. It means that there is no quantum state, where x or y has a definite value. The operators of x and y have no eigenstate. It seems that it all boils down to the continuous model of space and time. See this topic: A problem with the uncertainty principle.