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Why the tensor product space?

  1. May 2, 2013 #1
    Hi everyone,

    I'm reading through tensor product spaces and one question really bogs me. Why is it that the total Hilbert space of a system composed of two independent subsystems is the tensor product of the Hilbert spaces of the subsystems?

    It is always posed, but I've never seen a proof or a single argument for this. Sure, it works, but what is the motivation?

    Thanks for any help
     
  2. jcsd
  3. May 2, 2013 #2
    I think it is a postulate. It also gives the correct equations for joint probability distributions and is, in a sense, the natural way to represent a joint wavefunction.

    This might not be the best explanation (it is certainly not he most well put together), but here is another argument for it.
    Say you have two disconnected systems, and have observables A on the first and B on the second. It is natural to require that measuring both A and B "simultaneously", the joint observable AB is linear in both arguments: [tex]AB|\psi,\phi\rangle =A|\psi\rangle B|\phi\rangle[/tex] or in a more suggestive notation, [tex]A\otimes B(|\psi\rangle \otimes |\phi\rangle) =(A|\psi\rangle)\otimes (B|\phi\rangle)[/tex] where A and B are linear operators. In short, an observable on each of two disconnected systems gives rise to a unique observable on their tensor product.
     
  4. May 2, 2013 #3

    DarMM

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    It's basically espen180's argument. That argument shows that all states of the form:
    [tex]\Psi \otimes \Phi[/tex] should be states of the total system.

    Then, under the normal rules of quantum mechanics, any linear combination of such states should
    also be a state (principle of superposition) this gives you a space [tex]\mathcal{H}[/tex] Finally all Cauchy sequences in [tex]\mathcal{H}[/tex] should converge, since the limit of any Cauchy sequence will always correspond to the action of some unitary operator on a state already in [tex]\mathcal{H}[/tex] So adding in all such limits gives [tex]\overline{\mathcal{H}}[/tex] which is the tensor product of the two original spaces.
     
  5. May 2, 2013 #4

    Fredrik

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    This isn't a very rigorous argument, but at least it's an argument...

    Consider two systems that aren't interacting with each other. If system 1 is in state ##|\psi\rangle## when we measure A, the probability of result ##a## is
    $$P(a)=\left|\langle a|\psi\rangle\right|^2.$$ If system 2 is in state ##|\phi\rangle## when we measure B, the probability of result ##b## is
    $$P(b)=\left|\langle b|\phi\rangle\right|^2.$$ I'm not sure what the notational conventions are for tensor products of bras, so maybe a and b should be swapped on the right-hand side below... The standard rules for probabilities tell us that the probability of getting both of these results is
    $$P(a\ \&\ b)=P(a)P(b)=\left|\langle a|\psi\rangle\right|^2\left|\langle b|\phi\rangle\right|^2=\left|\langle a|\otimes\langle b|\ |\psi\rangle\otimes|\phi\rangle\right|^2.$$ So if we use the tensor product space to represent the states of the combined system, the Born rule will hold for that space too.
     
  6. May 2, 2013 #5

    DarMM

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    The only actual proof that I'm aware of uses the GNS theorem, where the use of the tensor product is just the noncommutative generalisation of the use of [tex]A \times B[/tex] as the sample space for a system composed of two independent subsystems in normal Kolmogorov probability.

    Many things that seem mysterious in quantum mechanics are really just noncommutative versions of things you already know from normal probability. A good book on this is Statistical Dynamics: A Stochastic Approach to Nonequilibrium Thermodynamics by Raymond Streater, see chapter 8.
     
  7. May 2, 2013 #6

    Fredrik

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    A small LaTeX tip for DarMM. You can use ## the way you would use a single dollar sign in a LaTeX document (i.e. instead of itex tags). You can also use $$ instead of tex tags. Hit the quote button next to post #4 to see examples.

    Regarding an actual proof...I think the 1978 article by Aerts & Daubechies gets the job done, but I haven't studied it in detail and it's been a couple of years since I looked at it, so I'm not sure I would be able to answer any questions about it.

    https://web.math.princeton.edu/~ingrid/publications/AD78.pdf
     
  8. May 3, 2013 #7
    Wow, thanks for the comments! I'll be sure to look into these arguments, thanks a lot!
     
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