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Conceptual question of many particle quantum systems and state vectors

  1. Jan 8, 2014 #1
    Apologies if this seems a bit trivial, but I was hoping to clear up a bit of confusion that I'm having.

    First Question: When considering an N-particle quantum system, described by some wave-function, [itex] \Psi\left(\vec{r}_{1},\ldots,\vec{r}_{N}\right)[/itex]. Is it correct to consider each particle to be in a given quantum state, such that the i-th particle to be in a given quantum state, at position, [itex] \vec{r}_{i}[/itex]? For the particular case of non-interacting particles, the wave-function can then be expressed as a linear combination of so-called 'single-particle states' ?
    (would it also be correct to say that a wave-function that can be expressed in terms of single-particle states contains the most 'complete' description one can obtain for a quantum system?)

    Second Question: Is it correct to consider a 'state-vector' (position representation), [itex] \lvert\Psi\left(\vec{r}\right)\rangle [/itex] as describing all possible states that the given quantum system (that it's describing) can be found in over all space?
    When we 'project' [itex] \lvert\Psi\left(\vec{r}\right)\rangle [/itex] onto one of the basis vectors, [itex] \lvert\vec{r}_{i}\rangle [/itex], is this the wave-function describing the state of system at position, [itex] \vec{r}_{i} [/itex]?

    Again, apologies in advance if this is complete rubbish, but would really appreciate if someone could shed some light on this for me.
     
  2. jcsd
  3. Jan 8, 2014 #2

    bhobba

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    No - because of the phenomena of entanglement. Suppose you can have particle 1 in the states with position x1 (state |x1>) and x2 (state |x2>) and similarly for particle 2. Particle 1 in state x1, and particle 2 in state x2 would be |x1>|x2>. Particle 1 in state x2, and particle 2 in state x1 would be |x2>|x1>. You can have a state that is a superposition of these ie c1*|x1>|x2> + c2*|x2>|x1>.

    What a state is, and how you should look at it, is very interpretation dependent. My view, purely on the basis of the formalism, without entering into the quagmire of these interpretational issues, is its simply what's required to allow us to predict the probabilities of outcomes of observations, similar to what probabilities themselves are. Indeed the differences between some interpretations is simply a variant of the different ways you can interpret probabilities ie frequentest or Bayesian, but I wont go into that right now.

    Here is a way of looking at it that clearly brings out whats going on with the formalism. I will base the following on the two axioms in Ballentine - QM - A Modern development. Its a very interesting fact that QM really rests on just two axioms. There is a bit more to it, but they are more or less along the lines of reasonableness assumptions such as the probability of outcomes should not dependent on coordinate systems ie symmetry.

    Imagine we have a system and some observational apparatus that has n possible outcomes associated with values yi. This immediately suggests a vector and to bring this out I will write it as Ʃ yi |bi>. Now we have a problem - the |bi> are freely chosen - they are simply man made things that follow from a theorem on vector spaces - fundamental physics can not depend on that. To get around it QM replaces the |bi> by |bi><bi| to give the operator Ʃ yi |bi><bi| - which is basis independent. In this way observations are associated with Hermitian operators. This is the first axiom in Ballentine, and heuristically why its reasonable.

    Next we have this wonderful theorem, Gleason's theorem, which, basically, follows from the above axiom:
    http://kof.physto.se/theses/helena-master.pdf [Broken]

    This is the second axioms in Ballentine's treatment.

    This means a state is simply a mathematical requirement to allow us to calculate expected values in QM. It may or may not be real - there is no way to tell. Its very similar to the role probabilities play in probability theory. In fact in QM you can also calculate probabilities. Most people would say probabilities don't exist in a real sense and why I personally don't think the state is real - but that's just my view - as far as we can tell today its an open question.

    Basically QM is a theory about the probabilities of outcomes of observation, if we were to observe it. The sole purpose of a state is, when combined with an observable, is to allow us to predict the probabilities of the outcomes of observations.

    This has nothing to do with what it MEANS. Different interpretations have different takes on that. But purely form the formalism that's what it's about.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  4. Jan 8, 2014 #3
    Thanks for the detailed description, much appreciated.
    Excluding entanglement though, would the description a gave (in the first part) be ok?
     
  5. Jan 8, 2014 #4

    bhobba

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    That's sort of a tautological statement - particles being in a definite state independent of the the state of other particles is the definition of not being entangled. Its true - but you are really saying the same thing.

    Thanks
    Bill
     
  6. Jan 8, 2014 #5

    atyy

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    No, because it isn't true even for the wave function of one particle.
     
  7. Jan 8, 2014 #6

    strangerep

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    Since no one seems to have addressed the OP's first question...

    No, multiparticle states are expressed in terms of tensor product spaces. E.g., a linear combination of products of single-particle states. Cf. Ballentine, ch17, p470.

    Not in interacting QFT in general. The interaction changes not only the Hamiltonian, but also the form of the single-particle states. But that's a long story. (Haag's Thm.)
     
  8. Jan 8, 2014 #7

    What about mutually interacting particles though, e.g. two electrons mutually repulsed due to like charge?


    Thanks strangerep, much appreciated.


    Is it true though to say that [itex]\Psi\left(\vec{r}_{1},\ldots,\vec{r}_{N}\right)[/itex] describes the state of the system, where [itex]\vec{r}_{1},\ldots,\vec{r}_{N}[/itex] are the coordinates of the N particles in the system?
     
    Last edited: Jan 8, 2014
  9. Jan 8, 2014 #8

    atyy

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    Yes.
     
  10. Jan 8, 2014 #9

    WannabeNewton

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  11. Jan 8, 2014 #10
  12. Jan 8, 2014 #11
    I think my confusion arises when discussing indistinguishable particles, namely when particle exchange is discussed. When we talk about a wave-function, [itex]\Psi\left(\vec{r}_{1},\ldots,\vec{r}_{i},\vec{r}_{j},\ldots,\vec{r}_{N}\right)[/itex] describing a particular micro-state of a quantum system and then exchange the positions of the i-th and j-th particles, such that the wave-function becomes [itex]\Psi\left(\vec{r}_{1},\ldots,\vec{r}_{j},\vec{r}_{i},\ldots,\vec{r}_{N}\right)[/itex], we say that at most they must differ by a phase, [itex]e^{i\alpha}[/itex], as there should be no detectable change in the observable state of the system. My confusion arises in whether, in the case of distinguishable particles (such that the exchange of two particles results in a different micro-state of the system), we can talk about the particles at positions [itex]\vec{r}_{i}[/itex] and [itex]\vec{r}_{j}[/itex] changing states, or whether we can only talk about the micro-state of the system changing as a whole?
     
  13. Jan 8, 2014 #12

    atyy

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    Even for one particle, ψ(x1) is not the state of the particle at x1, because the particle is not at x1. ψ(x1) is the state of particle 1.

    For two particles, ψ(x1,x2) is the state of the system. If the state can be written as ψ(x1,x2)=β(x1)γ(x2), then you could conceivably say that β is the state of particle 1 and γ is the state of particle 2. However, if the state can only be written as ψ(x1,x2)=βi(x1i(x2)+βj(x1j(x2), then one cannot say what "the" state of particle 1 is, because it is "correlated" with the state of particle 2.
     
    Last edited: Jan 8, 2014
  14. Jan 8, 2014 #13
    Is it correct though, in the case of distinguishable particles, to say that exchanging the coordinates of the i-th and j-th particles, such that [itex]\Psi\left(\vec{r}_{1},\ldots,\vec{r}_{i},\vec{r}_{j},\ldots,\vec{r}_{N}\right)\rightarrow\Psi\left(\vec{r}_{1},\ldots,\vec{r}_{j},\vec{r}_{i},\ldots,\vec{r}_{N}\right)[/itex] changes the state of the the system?

    Also, going back to an earlier question, would it be correct to say that [itex]\langle x\rvert\Psi\rangle=\Psi\left(x\right)[/itex] is the representation of the state of a given system in position space?
     
  15. Jan 8, 2014 #14

    atyy

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    Yes. You can see it concretely with ##\psi(x,y)=\alpha(x)\beta(y)##, where ##x## and ##y## are coordinates for particles 1 and 2 respectively. Then ##\psi(y,x)=\alpha(y)\beta(x) = \beta(x)\alpha(y)## which is a different wave function.

    If you have a system of two particles (particle 1 and particle 2), but you are somehow able to make a measurement of only particle 1, the "effective state" of particle 1 is usually obtained by writing the state of the combined system in the "density matrix" formalism, and doing a partial trace over particle 2.

    Eg:
    p4 of http://www.unc.edu/~marzuola/Math547_S13/Math547_S13_Projects/E_Thiede_Section003_DensityMatrices.pdf [Broken]
    p18 of http://www.pa.msu.edu/~mmoore/Lect34_DensityOperator.pdf
     
    Last edited by a moderator: May 6, 2017
  16. Jan 8, 2014 #15
     
  17. Jan 8, 2014 #16

    atyy

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    Yes, unless it is a state of some very special symmetry. For example, even for product states, if the wave function is ##\psi(y,x)=\alpha(y)\alpha(x)## then swapping ##x## and ##y## makes no difference.

    You can usually get the answers by trying simple examples for two particles.
     
  18. Jan 8, 2014 #17
    Ok, cheers :smile:
     
  19. Jan 9, 2014 #18

    bhobba

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    Particles are either created entangled or, via some kind of interaction become entangled. In fact, the phenomena of decoherence, which many like me think has something important to say about the measurement problem, is a result of interaction between the system, environment, and measuring apparatus (if any) and is a form of entanglement.

    There is view emerging, that I personally hold to, that the rock bottom essence of QM is entanglement:
    http://arxiv.org/abs/0911.0695

    Thanks
    Bill
     
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