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Dirac notation Schwarz Inequality Proof

  1. Jun 20, 2014 #1
    1. The problem statement, all variables and given/known data
    This isn't really a problem so much as me not being able to see how a proof has proceeded. I've only just today learned about Dirac notation so I'm not too good at actually working with it. The proof in the book is:

    |Z> = |V> - <W|V>/|W|^2|W>

    <Z|Z> = <V - ( <W|V>/|W|^2 ) W| V - ( <W|V>/|W|^2 ) W> =

    <V|V> - ( <W|V><V|W>/|W|^2 ) - ( <W|V>*<W|V>/|W|^2 ) + ( <W|V>*<W|V><W|W>/|W|^4| ) >= 0

    This last step has me completely lost, and if someone could explain it I'd really appreciate it. It looks like he's pulled out the <V| and |V> and combined them to get <V|V>, but I don't get where the minus sign comes from, or how he gets any of the rest of the equation, so I'm assuming I'm wrong about where the <V|V> comes from.

    I'm deeply sorry about the notation here, I typed this on a phone or I would've used LaTeX.


    2. Relevant equations

    <Z|Z> >= 0

    3. The attempt at a solution

    Mentioned in the first section.
     
  2. jcsd
  3. Jun 20, 2014 #2

    kreil

    User Avatar
    Gold Member

    So you know that

    [tex]
    |Z\rangle = |V\rangle - \frac{\langle W|V\rangle}{|W|^2} |W\rangle
    [/tex]

    So what do you think [itex]\langle Z|[/itex] equals?
     
  4. Jun 21, 2014 #3
    [itex]\langle Z|[/itex] should just equal [tex]
    \langle V| - \frac{\langle W|V\rangle}{|W|^2} \langle W|
    [/tex]

    right? So if I look at [itex]\langle Z|Z \rangle[/itex], I have to distribute, but I can't see how he's getting the terms he's getting, namely because I don't think I'm too comfortable with Dirac notation yet.

    1: [itex]\langle V | V \rangle = \langle V | V \rangle[/itex] obviously,

    2: [itex]\langle V | - \frac{\langle W|V\rangle}{|W|^2} W \rangle[/itex] seems like it should equal [itex]- \frac{\langle W|V\rangle \langle V|W \rangle}{|W|^2}[/itex]

    3: Probably the most confusing term for me:

    [itex]\langle - \frac{\langle V|W \rangle}{|W|^2}W|V \rangle[/itex]

    This seems to me like it should just be [itex]- \frac{\langle V|W \rangle \langle W|V \rangle}{|W|^2}[/itex] but this term doesn't appear anywhere in his final formulation really. I mean I guess [itex]\langle V|W \rangle[/itex] is equivalent to his [itex]\langle W|V \rangle ^*[/itex], but why does he get that conjugate instead of the [itex]\langle V|W \rangle[/itex] that I got?

    4: then [itex]\langle - \frac{\langle W|V\rangle}{|W|^2} W|- \frac{\langle W|V\rangle}{|W|^2} W \rangle[/itex] is the first term where I think I'm getting confused. For some reason the term he gets from this has a complex conjugate in it, which I assume means he's moving [itex]| - \frac{\langle W | V \rangle}{|W|^2} \rangle[/itex] to the bra side, which I assume makes the top term a complex conjugate, leaving me with [itex]\langle - \frac{\langle W|V\rangle}{|W|^2}\frac{\langle W|V\rangle ^*}{|W|^2} W| W \rangle = \frac{ \langle W|V \rangle \langle W|V \rangle ^* \langle W|W \rangle}{|W|^4}[/itex]

    At the end, my final answer would look like: [itex]\langle V | V \rangle - \frac{\langle W|V\rangle \langle V|W \rangle}{|W|^2} - \frac{\langle V|W \rangle \langle W|V \rangle}{|W|^2} + \frac{ \langle W|V \rangle \langle W|V \rangle ^* \langle W|W \rangle}{|W|^4}[/itex]

    This seems like it may at least be equivalent to his formulation based on what I said in 3, but even this took kind of a mental leap for me in part 4 with moving a term over to the bra side. Is that a correct step? If so, why?

    Thanks so much for any help. I'm losing sleep over this.
     
  5. Jun 21, 2014 #4

    kreil

    User Avatar
    Gold Member

    No problem, anyone that gets a science degree loses some sleep over a problem or two, so I know what boat you're in!

    Does it help if I say that,

    [itex]
    |Z\rangle = \langle Z|^{\dagger}
    [/itex]
    [itex]
    \langle Z| = |Z\rangle ^{\dagger}
    [/itex]

    and that,

    [itex]
    \langle W |V \rangle ^{\dagger} = \langle W | V \rangle ^* = \langle V|W\rangle ...?
    [/itex]

    (In other words, the inner product of a bra and a ket is a complex number, and the Hermitian conjugate of a complex number is simply its complex conjugate.)

    This all means that,

    [tex]
    \langle Z| = |Z\rangle^{\dagger} = \left( |V\rangle - \frac{\langle W|V\rangle}{|W|^2} |W\rangle \right) ^{\dagger} =\left( \langle V | - \frac{\langle W|V \rangle ^*}{|W|^2} \langle W| \right)
    [/tex]

    So,

    [tex]
    \langle Z | Z \rangle = \left( \langle V | - \frac{\langle W|V \rangle ^*}{|W|^2} \langle W| \right) \left( |V\rangle - \frac{\langle W|V\rangle}{|W|^2} |W\rangle \right)
    [/tex]
    See also,
    http://en.wikipedia.org/wiki/Bra–ket_notation#Inner_products_and_bras
     
  6. Jun 21, 2014 #5
    Yes, that clears it up completely, thank you.

    I can now actually see the part in the book where the asterisk denoting that the initial setup involved a complex conjugate has been worn away due to shoddy printing, so I feel a lot better about this whole thing in general now.

    Thanks so much!
     
  7. Oct 4, 2015 #6
    Hmm.. I don't get why the proof starts with that |Z>
    Mind to explain it, please? thanks :)
     
  8. Dec 17, 2016 #7
    Towards the end they consider, <W|V><V|W> = |<V|W>|, which I cannot understand. Please explain someone.
     
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