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Questions about fundamental concepts

  1. Jul 2, 2013 #1
    I am confused about what I think should be the most basic concepts in quantum mechanics..

    As far as I understood, a system that can be in only one of two states, can literally be only in one of those two states, right? So if I were to make a hypothetical measurement of the configuration of that system, I could only get one of those states as a result of that measurement (and never something in between, mostly because there is nothing in between).

    But then there is the idea of the superposition of those states. Which implies that my system can abstractly be in one of these non-existent states. I will never get one of those as a result of an experiment, but it reflects my uncertainty of the state of that system in a particular instant in time, right?

    I guess my confusion starts with the motivation of inventing these in-between states. Is it because my system's time evolution is probabilistic? So even though I may choose to start it at some realizable state, in a couple of seconds I might only be able to predict the state of that system up to a probabilistic result?

    So would the bottom line be: in classical mechanics I can just say that in three seconds from now your system is going to be in the state X. In quantum mechanics I don't want to say "I don't know", so I just say in three seconds from now your system is going to be in the "state" of having 30% probability of being in state X and 70% probability of being in state Y?

    Than my final question is why do we use probability amplitudes to talk about these states rather than just straight up probability?
  2. jcsd
  3. Jul 2, 2013 #2
    Superposition implies it's in a linear combination of the two states.

    A system can be in the superposition of two states without time evolution. For example, a photon polarized at a 45 degree angle. It is half vertical and half horizontal. If it hits a sheet that classically only lets through vertical photons, it will transmit with 50% probability. Superposition isn't about uncertainty. It's not that the particle has a 50% chance of being in one of the other, and we don't know until we check. The particle is really in BOTH until we check. Even the particle doesn't know which state it's in until something that checks whether it's in one or the other forces it to choose.

    Now I think what you were describing is wave packet spread which does involve time evolution. It's easiest to understand it when thinking about the free particle. In CM, given postion and velocity, you can predict the trajectory of the particle. But QM, there are uncertainty relations that need to be satisfied and you can't be sure how exactly the state of the particle with evolve (since you don't exactly know both the position and momentum), but you can tell what range it will be in. As time evolves the range will grow larger and larger until you make a measurement and collapse the wavefunction.

    We use probability amplitudes because two states can differ by a phase, but in the end yield the same probabilities. The phase difference could be important when calculating the projection of a state onto another state.

    Hopefully that clears some things up, but maybe I used too much jargon. Maybe I could have arranged the paragraphs differently...
  4. Jul 3, 2013 #3


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    It's more subtle than that. The reason is that for any pure quantum state (at least in finite dimensions) then at least in principle there's always a measurement that could be performed on it that would produce a deterministic outcome.

    Example: suppose [itex]\lvert 0 \rangle[/itex] and [itex]\lvert 1 \rangle[/itex] are two perfectly distinguishable states in some two level system (e.g. two atomic levels, spin up/spin down, ...). According to quantum physics, the system is allowed to be in the superposition state
    [tex]\lvert + \rangle = \frac{1}{\sqrt{2}} \bigl( \lvert 0 \rangle + \lvert 1 \rangle \bigr) \,.[/tex]
    If you perform a measurement to try to distinguish between the [itex]\lvert 0 \rangle[/itex] and [itex]\lvert 1 \rangle[/itex] states, you'll get the result [itex]\lvert 0 \rangle[/itex] with probability 1/2 and the result [itex]\lvert 1 \rangle[/itex] with probability 1/2.

    But according to quantum physics, you could just as well decide you want to do a measurement that distinguishes between the states [itex]\lvert + \rangle[/itex] and [itex]\lvert - \rangle[/itex], where
    [tex]\lvert - \rangle = \frac{1}{\sqrt{2}} \bigl( \lvert 0 \rangle - \lvert 1 \rangle \bigr) \,.[/tex]
    If you do this measurement on the [itex]\lvert + \rangle[/itex] state, then in this case you'll detect the [itex]\lvert + \rangle[/itex] state with certainty.

    (If you think of [itex]\lvert 0 \rangle[/itex] and [itex]\lvert 1 \rangle[/itex] as "spin up" and "spin down", then [itex]\lvert + \rangle[/itex] and [itex]\lvert - \rangle[/itex] might correspond to "spin left" and "spin right". Measuring [itex]\{ \lvert 0 \rangle, \lvert 1 \rangle \}[/itex] or [itex]\{ \lvert + \rangle, \lvert - \rangle \}[/itex] correspond to measuring the spin projection along the z or x axes, respectively.)

    If you're in a situation where a system might be in the state [itex]\lvert 0 \rangle[/itex] or in the state [itex]\lvert 1 \rangle[/itex] and you simply don't know which, it's actually wrong to describe that with a superposition like [itex]\tfrac{1}{\sqrt{2}} \bigl( \lvert 0 \rangle + \lvert 1 \rangle \bigr)[/itex], because if you write something like that you're implying you know much more than you actually do about the possible results of different possible measurements.
  5. Jul 3, 2013 #4
    What does it mean to make a measurement of the ## \lvert + \rangle ## state? And also, how can you detect the ## \lvert + \rangle ## state? If my system has two states, ## \lvert 1 \rangle ## and ## \lvert 0 \rangle ##, what did I just detect?

    So there no such thing as a quantum system being in one state? It's always in every realizable state it could be? And we represent that using some superposition of those realizable states? I mean, there is no ambiguity as to what state [itex] \lvert A \rangle = \alpha\lvert\ 1 \rangle + \beta\lvert\ 0 \rangle [/itex] my system is in. If it is in that state, it is in THAT state and no other? But as far as being in either ## \lvert 1 \rangle ## or ## \lvert 0 \rangle ## I don't know. It's just in the state ## \lvert A \rangle ##...?

    So what does it mean to take the amplitude (bracket) between two arbitrary states? Is that like what's the probability amplitude for being in both states at the same time? or the probability amplitude for being that I am in one of those states how likely was I to have been in the other?

    I am still confused..

    I guess I can pin down my confusion in what the probability amplitude measures.
  6. Jul 3, 2013 #5


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    Staff: Mentor

    The states are vectors in an abstract vector space, so a state can be written as different linear combinations of different basis vectors yet still be the same state. It's analogous to the way that a vector of magnitude A pointing to the northeast can be written as ##A\hat{r}## where ##\hat{r}## is a unit vector pointing northeast or as ##\frac{A\sqrt{2}}{2}(\hat{n}+\hat{e})## where ##\hat{n}## and ##\hat{e}## are unit vectors pointing north and east. It's the same vector either way, and I can use whichever form is more convenient for the problem at hand.

    If ##\vert{a_1}\rangle## and ##\vert{a_2}\rangle## are the eigenfunctions of operator A with eigenvalues ##a_1## and ##a_2##, ##\vert{b_1}\rangle## and ##\vert{b_2}\rangle## are the eigenfunctions of operator B with eigenvalues ##b_1## and ##b_2##, and A and B don't commute, then I can write a state ##\vert\psi\rangle## as:
    [tex]\vert\psi\rangle = c_1\vert{a_1}\rangle + c_2\vert{a_2}\rangle = c_3\vert{b_1}\rangle + c_4\vert{b_2}[/tex]

    This tells me that if I measure A, I will get result ##a_1## with probability ##{c_1}^2## and ##a_2## with probability ##{c_2}^2##, and similarly if I measure B I'll get ##b_1## with probability ##{c_3}^2## and ##b_2## with probability ##{c_4}^2##. The first representation is more useful if I want to know about the value of A and the second if I want to know about the value of B... but it's all the same state, just different mathematical representations.

    Does this mean the system is simultaneously ##a_1## and ##a_2## and ##b_1## and ##b_2## until I've measured something? I don't think that's a helpful way of describing what the mathematical formalism is telling us.
  7. Jul 3, 2013 #6
    It may be that ##|+\rangle## is not a state that we have any mechanism for measuring. It depends on particular system. Think of this the way classical mechanics works: nobody ever thought that we could perfectly measure the position of a particle, they just assumed that it was the fault of our measuring devices. Similarly, the state ##|+\rangle## is a valid state that could be (theoretically) measured, but we may not have the appropriate technology.

    An example of observables that work like this is spin where the pure state ##X## is a superposition of ##Y## and ##Z##.

    First of all, the math in QM can sometimes have multiple meanings depending on the situation. Most of the time, ##\langle A|B\rangle## is interpreted as the probability that the state ##|A\rangle## will be measured assuming that the state was originally prepared in ##|B\rangle##. Was that the question you were asking? Are you taking a QM class or self studying from a book? The sorts of things are best understood by following along with exercises and working easy problems.
  8. Jul 3, 2013 #7
    All the responses here are very good conceptual examples to keep in mind when thinking about QM. But I think a less mathematical example might help you to actually get a feel for what's going on.

    I think the best way to think of a superposition of states is to remember the single-particle double-slit interference pattern. Here's a nice video of the experiment actually being performed: http://youtu.be/jvO0P5-SMxk . The take-away message from this experiment is that when a single particle (e.g. an electron) is shot such that it can pass through two slits, with equal probability for passing through each slit, then the resulting trajectories the particle takes show an interference pattern between the two possible paths. It's not just a mathematical interpretation--if there's only one particle, but the particle is interfering with something, then it must be interfering with itself. If it weren't simultaneously in two states at once, we couldn't get the interference pattern that develops in the experiment in the video. So superpositions of states DO happen and experiments can't be explained without them.

    On the other hand, in that same experiment, if one were to take measurements right inside the slits to determine which slit the particle actually passes through, then we get an entirely different result because the superposition is lost and thus the particle cannot interfere with itself. So the act of measuring actually changes whether the particle is superposed or not.

    This all traces back to the fact that in Quantum Mechanics, there are only two possible time-evolution processes: 1) Schrodinger Evolution and 2) Wavefunction Collapse.

    Schrodinger evolution is what happens between measurements. An electron initially located in one spot can evolve according to Schrodinger evolution into a state where it is simultaneously passing through two different slits. (Schrodinger evolution is entirely deterministic and there is no uncertainty in how a quantum system evolves when it is not being measured.)

    Wavefunction collapse happens when you make a measurement. Whatever superposed state you choose to measure somehow instantly turns into an "eigenstate" or a normal state where the particle is not doing two things at once--this instantaneous change in state is called wavefunction collapse. (The reason things are probabilistic is because the wavefunction collapses into one of the constituent states that make up the superposed wavefunction, but there is uncertainty as to which one it actually collapses into.)

    So the reason we always observe things to be in their normal eigenstates is because our measurement forces it to be that way. However, when left to its own devices, a quantum state will generally evolve into some superposition state--and this superposition is actually a real thing that must be there in order to make sense of actual experiments.

    One thing I should mention though is that adherents to certain interpretations would argue that Wavefunction collapse never actually happens; only Schrodinger evolution occurs, and the forcing of things into eigenstates happens as a result of "decoherence" which reproduces in every way the appearance of collapse.

    Edit: Wiki actually has a nice section on what I was talking about with regards to the single-particle double-slit experiment. http://en.wikipedia.org/wiki/Double_slit#Variations_of_the_experiment
    Last edited: Jul 3, 2013
  9. Jul 3, 2013 #8
    I just realized that what was really confusing me is that there are different interpretations to quantum mechanics. I though I didn't understand how everything fit together, but in reality it doesn't. And now everything makes a lot of sense!

    Thanks! :)
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