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Order of operations with bras and kets

  1. Sep 12, 2010 #1
    I am confused about the order of operations with kets, as well as when one can permute the order of the kets.

    Question 1:
    When I have a term as follows:

    ...why exactly am I able to switch the order of the bras and kets as follows?

    I get confused about when we are able to change the order of the bras/kets and when we have to preserve their order.

    Question 2: And why don't we look at the previous state and say that <+|d+> = 0? In that case the entire term would be zero, which is obviously not the case. I get that the states can't be called orthogonal because they refer to different objects, and hence to different Hilbert spaces...is that the correct answer? What, then, is <+|d+> equal to?

    Those are my questions. For the setup to this question, read on...This example is taken from Zurek 1991 (Decoherence and the Transition from Quantum to Classical).

    A particle in state
    |K> = a|+> + b|->

    interacts with a quantum detector/system, and becomes

    |K>= a|+>|d+> + b|->|d->

    where |d+> and |d-> stand for the up and down states of the quantum detector. (we assume the detector is not macroscopic, so we avoid any confusion about the existence of Macroscopic Quantum States).

    When I create the density matrix for this state, I get
    |K><K| = a^2|+>|d+><+|<d+| + b^2|->|d-><-|<d-| +....plus a couple other off-diagonal terms.
  2. jcsd
  3. Sep 12, 2010 #2
    Perhaps your confusion comes from the fact that you are not switching the order. If, for instance:

    [tex]|K>= a|+>|d+> + b|->|d->[/tex]


    [tex]<K| =a^*<d+|<+|+b^* <d-|<-|[/tex]
  4. Sep 12, 2010 #3
    Thank you, that is helpful.

    So, then, if we form the projector:

    |K><K| = a^2|+>|d+><d+|<+| +....other terms

    (I have properly written the complex conjugate in reverse order)

    ...how do we move the <+| so it is next to the |+>? Clearly this is some simple linear algebra step I am missing.
  5. Sep 12, 2010 #4
    Whenever you see ....<x1|x2>.... , where x1 and x2 refer to one system - that is a c-number. So you move it in front of the expression and analyze what remains.
    Whenever you see ....|x><x| - this is a projection operator. Projection operators belonging to one system commute with the variables of the other system - they do not affect them. This applies also to more general operators of the form |x1><x2|.
    Last edited: Sep 12, 2010
  6. Sep 12, 2010 #5
    The kets [itex]|+\rangle[/itex] and [itex]|d+\rangle[/itex] refer to different systems, so the product [itex]|+\rangle |d+\rangle[/itex] is a tensor product, it could be written [itex]|+\rangle \otimes |d+\rangle[/itex]. Sometimes you will see [itex]|+d+\rangle[/itex] meaning the same thing.

    The outer product of tensor products [itex](|+\rangle \otimes |d+\rangle )( \langle +| \otimes \langle d+ |)[/itex] can be written as the tensor product of outer products [itex](|+\rangle \langle +| )\otimes (|d+\rangle \langle d+ |)[/itex]. This is just a property of the tensor product.

    The inner product [itex]\langle +|d+\rangle[/itex] isn't defined since it's between different Hilbert states. I had a look at the Zurek paper (the version on arxiv, http://arxiv.org/ftp/quant-ph/papers/0306/0306072.pdf" [Broken]) you mentioned, he has two |s next to each other like this [itex]|+\rangle \langle +||d+\rangle \langle d+ |[/itex] instead of one like you do in an inner product, except once in eqn 6, which is a typo. People often omit the [itex]\otimes[/itex] symbol for some reason.
    Last edited by a moderator: May 4, 2017
  7. Sep 12, 2010 #6


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    You're not. There must be a [itex]\otimes[/itex] missing, as Tomsk said. There should be a couple of [itex]\otimes[/itex] symbols in the first expression as well, but leaving them out there isn't as disastrous as in the second expression. Insert the [itex]\otimes[/itex] symbols, and then prove equality by having both of these operators act on the same arbitrary vector [itex]|\alpha\rangle\otimes|\beta\rangle[/itex].
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