0

If you look at the commutation relation of the position and momentum operators (in 1D position space), you get:

$$[\hat{x}, \hat{p}_x] = [x,-i \hbar \frac{\partial}{\partial x}] = i \hbar$$

All this says to me is that if you prepare a system in state (A) and measure the position, the system is now in state (B), which is an eigenstate of the position operator. You then measure the momentum of state (B) and now you're in state (C), which is an eigenstate of the momentum operator. Note that these must be consecutive measurements.

Alternatively, let's assume you reversed the order of the measurements. Beginning with state (A) again, you measure momentum first and put the system in state (D), an eigenstate of the momentum operator, but not necessarily state (C) [right?]. You then measure position, and put the system in state (E).

$$(B) \neq (E)$$ $$(C) \neq (D)$$

That's all HUP says to me. It says nothing about simultaneous measurements of position and momentum — is that even possible? (What operator would that be?) It only says how much the final wavefunctions of two systems that started out the same differ if you perform two measurements in a different order.

Where is the uncertainty? You know the position and momentum exactly — right when you measure them. You get some random value weighted by the coefficients of the eigenfunctions in the linear combination that forms $\psi$.

So I don't think it's accurate to say you can't "simultaneously know position and momentum to arbitrary precision", because as far as I can tell, you can't even measure the two at the same time.

Nick
  • 2,949

2 Answers2

2

Most of your question is phrased in terms of measurement, but in the final paragraph you talk about what there is in principle to be known. These are two different interpretations of the HUP. The measurement interpretation is the original one found in Heisenberg's 1927 paper, but today we really think of it in terms of what there is to be known.

There is some discussion of this point in this answer by David Z, and also in the introduction to the WP article. Distler 2012 shows that the measurement interpretation has some fundamental problems:

[...] the Uncertainty Principle does not say anything about successive measurements. In fact, formulating a precise statement about the uncertainties in the result of successive measurements (of non-commuting observables) has received, perhaps, less attention than it deserves.

Nick wrote:

So I don't think it's accurate to say you can't "simultaneously know position and momentum to arbitrary precision", because as far as I can tell, you can't even measure the two at the same time.

This can't be just trivially true, independent of the commutation relation between the operators, because you can simultaneously know, e.g., the angular momentum and parity of a $^4\text{He}$ nucleus that's in its ground state -- the state is $J^\pi=0^+$.

Knowing two facts at the same time doesn't require observing them both at the same time. It only requires that neither fact has changed since the time when you measured it.

References

Distler and Paban, Uncertainties in Successive Measurements, http://arxiv.org/abs/1211.4169

Community
  • 1
  • "today we really think of it in terms of what there is to be known" That doesn't sound quite right. QM consists of a PDE and a wavefunction collapse law (or some other related notion) without making any reference to a notion of knowledge. HUP is a mathematical statement about wavefunctions. – Dan Piponi Sep 24 '13 at 20:36
  • @DanPiponi: I disagree. I'll edit the answer to provide a reference and more discussion of this point. –  Sep 24 '13 at 23:21
  • Well I 100% agree with the point that HUP isn't about successive measurements. In fact, if you read what Heisenberg wrote on this subject he was himself confused about this point. But HUP isn't about knowledge either because knowledge isn't part of the vocabulary of QM. HUP is about wavefunctions. – Dan Piponi Sep 24 '13 at 23:42
  • 1
    @DanPiponi: The HUP is about wavefunctions, or more pedantically the state which wf is a representation of, but that this describes "what there is to be known" is just another way of saying that QM's description of the observables under consideration is complete, i.e., there is nothing further that determines them that we're just failing to measure. To me it is looks to be clearly what Ben is talking about, and is furthermore the standard view of QM (up to some commonly held very general assumptions that aren't very relevant here), your criticism seems a bit too nitpicky. – Stan Liou Sep 25 '13 at 00:38
2

All this says to me is that if you prepare a system in state (A) and measure the position, the system is now in state (B), which is an eigenstate of the position operator.

[This is less an answer and more a comment on the above but it's too long for a comment.]

But that's not what the commutation relation "says". Simply operating on a state with the position or momentum operator does not "collapse" the state to an eigenstate. For example, let:

$\lvert \psi \rangle = \alpha\lvert p_1 \rangle + \beta\lvert p_2 \rangle$

be a superposition of two momentum eigenstates. Then:

$\hat P \lvert \psi \rangle = \alpha p_1\lvert p_1 \rangle + \beta p_2 \lvert p_2 \rangle$

which is not a momentum eigenstate.

It is the projection postulate that "says" that a measurement of an observable will be an eigenvalue of the associated operator and that, immediately after the measurement, the state will be the associated eigenstate.

Alfred Centauri
  • 59,560