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I am trying to learn more about quantum mechanics. I am reading a book by Griffiths that I like. I'm trying to summarize what I've learned. So below I provided three assumptions. I'd like to know if they are correct.

Consider a particle in space and time.

  1. We cannot know where the particle is with certainty. If we perform the same measurement experiment on an ensemble of identically prepared quantum systems, on average we may find the particle at one location more often than others.
  2. Under certain conditions, we can know and predict future probability distributions. Specifically, this is when the probability of the particle distribution is constant in time (eg stationary states).
  3. Given a probability amplitude $\Psi(x,0)$, we can predict the future value $\Psi(x,t)$ through the Schrödinger equation.

I'm pretty sure I'm misunderstanding something. In particular, No.3 suggest that I could predict future probability distributions of the wave function and I know from talking to people on SE that's wrong.

My Question:

Can someone explain which (if any) of my assumptions above is wrong and explain why?

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    $\begingroup$ Nothing wrong to see here, move along. Who says we cannot predict future probability distributions? (See, for example, this question where all pretty much agree that the evolution equation for a quantum state is deterministic.) $\endgroup$
    – ACuriousMind ♦
    Commented Feb 20, 2015 at 2:13
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    $\begingroup$ Seems like you've been talking to some weird people on SE. $\endgroup$
    – zzz
    Commented Feb 20, 2015 at 2:24
  • $\begingroup$ Or perhaps I am just an unreliable interpreter of what they have said. lol mistakes and learning tend to go together. $\endgroup$
    – Stan Shunpike
    Commented Feb 20, 2015 at 3:13

2 Answers 2

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We cannot know where the particle is with certainty.

The particle, in general, does not have a definite location to know.

Under certain conditions, we can know and predict future probability distributions.

The evolution of the state is determined by the Hamiltonian (in the Schrodinger picture). The problem is that we don't know the Hamiltonian for the measurement apparatus. Thus, the certain conditions are that the Hamiltonian is known. The "collapse of the wavefunction" is essentially a reflection or our ignorance of the Hamiltonian including the measurement apparatus.

Given a probability amplitude Ψ(x,0), we can predict the future value Ψ(x,t) through the Schrödinger equation.

Correct in principle. The devil is in the details of the time evolution operator.

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  • $\begingroup$ I see. So your version of my (1) is probably more correct. Although "we cannot know it" is a consequence of the particle doesn't have a position to know, my statement makes it sound as if the problem could be with the measurement process itself. $\endgroup$
    – Stan Shunpike
    Commented Feb 20, 2015 at 3:11
  • $\begingroup$ Regarding point 2, what are we defining as the "measurement apparatus"? How do we know if we do or do not have a Hamiltonian for it? $\endgroup$
    – Stan Shunpike
    Commented Feb 20, 2015 at 3:12
  • $\begingroup$ @StanShunpike, measurement is inescapably an interaction. Either that measurement interaction is included in the Hamiltonian (which governs the time evolution) or it isn't. $\endgroup$
    – Alfred Centauri
    Commented Feb 20, 2015 at 3:16
  • $\begingroup$ @StanShunpike, regarding your first comment, on the Bohmian interpretation, the quantum particle (beable) has, at all times, a definite position and velocity. The uncertainty is rooted in our inescapable ignorance of the initial configuration. However, on the 'orthodox' view, the uncertainty is a feature of reality, i.e., the quantum 'particle' does not, in fact, have a definite position or whatever in the instant before a measurement is made. $\endgroup$
    – Alfred Centauri
    Commented Feb 20, 2015 at 3:22
  • $\begingroup$ so when you say "The "collapse of the wavefunction" is essentially a reflection or our ignorance of the Hamiltonian including the measurement apparatus." do you mean that....we don't how to formulate this Hamiltonian...or it doesn't exist...or both? $\endgroup$
    – Stan Shunpike
    Commented Feb 20, 2015 at 3:31
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  1. Correct, expect just don't say "on average," just we will find the particle at different locations more often than others in accordance with the probability distribution described by a wave function.

  2. Incorrect, you may always calculate the evolution of an initial state if you know the effective Hamiltonian or effective Action the state is subject to. However, it may not be easy or even possible to solve the quantum system analytically i.e. with out the use of an approximation or computer. The reason you see stationary states is because they tend to be easier to solve in closed form, hence the simplest of quantum systems appear in introductory texts.

  3. Correct.

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  • $\begingroup$ Ah, perhaps that's the confusion I have. Point two. What does it mean to solve it analytically? $\endgroup$
    – Stan Shunpike
    Commented Feb 20, 2015 at 2:17
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    $\begingroup$ To be able to express the answers in a form that uses algebraic or transcendental functions (sin, cos, exp, ...), for you this means being able to write the answer on paper. For example how might you solve the for the wave function of a really complex molecule? Or even an atom who has more than one electron? You can write down the Schrödinger equation but it will be very difficult if not even possible to solve it exactly (without approximation). $\endgroup$
    – BVPhD
    Commented Feb 20, 2015 at 2:19

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