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Shankar Questions About Quantum Mechanics (Schwarz Inequality)

  1. Dec 24, 2012 #1
    Hello there,

    Im studying QM with Shankar's book.

    I'm wrestling myself trough the linear algebra now and I have some questiosn.

    Let me start with this one:

    aajHi.jpg

    I have absolutely no idea where this is coming from or what does it mean.

    I don't know how to multiply a ket with an inner product...

    Thx in advance
     
  2. jcsd
  3. Dec 24, 2012 #2

    vanhees71

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    Re: Shankar Questions About Quantum Mechanics

    This is the completeness relation for a bunch of vectors, usually a complete orthonormalized set of Hilbert-space basis vectors, [itex]|i\rangle[/itex]. These usually occur as eigenvectors of an self-adjoint operator.

    Usually you need a slight extension of this concept, namely the case for unbounded essentially self-adjoint operators which can have a continuous spectrum (like the position "eigenkets", which are no Hilbert-space vectors but belong to a larger space, i.e., the dual space of the domain of the position and momentum operator). In this case your sum goes over into an integral
    [tex]|V \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; |\vec{x} \rangle \langle \vec{x}|V \rangle.[/tex]
    Sometimes also the case occurs, where an operator has both a discrete and a continuous spectrum, e.g., the Hamiltonian for the motion of a particle in the Coulomb potential of another heavy charged particle.
     
  4. Dec 24, 2012 #3
    Re: Shankar Questions About Quantum Mechanics

    To expand on vanhees's statement a bit, when we say "completeness," we are talking about a basis (usually an eigenbasis of a self-adjoint operator) for a particular vector space V. If we have a complete basis for a particular space (in this case, [itex]|i\rangle[/itex] collectively spans the vector space V), then we can express any arbitrary vector v (or function f if we're talking about a function space) as a sum of its components multiplied by the basis vectors. For example, if we're working in [itex]\Re^3[/itex], we can choose as a basis the standard basis vectors: [itex]e_1=(1,0,0)[/itex], [itex]e_2=(0,1,0)[/itex], and [itex]e_3=(0,0,1)[/itex]. In Dirac notation, these would typically be [itex]|1\rangle[/itex], [itex]|2\rangle[/itex], [itex]|3\rangle[/itex].

    Now given an arbitrary vector in [itex]\Re^3[/itex], we can write it as a sum of [itex]|1\rangle, |2\rangle, |3\rangle[/itex] with the proper coefficients in front of each basis vector. For example, the vector [itex]u=(2,1,0)[/itex] can be expanded in the form [itex]|u\rangle = \sum_{i=1}^{3} | i \rangle \langle i | u \rangle [/itex] as [itex]| u \rangle = 2 | 1 \rangle + 1 | 2 \rangle + 0 | 3 \rangle [/itex].

    The term [itex]\langle i | u \rangle [/itex] in the sum is the inner product of the basis vector with the arbitrary vector u. This inner product picks out the component of u in the direction of that particular i basis vector. Now this inner product is just a scalar value. The additional [itex]| i \rangle[/itex] is there because we're attempting to expand a vector as a sum of basis vectors multiplied by their respective components in u for each basis vector. So when you multiply an inner product by a ket, you're just scaling a ket. That's all.

    I hope this helps!
     
  5. Dec 24, 2012 #4
    Re: Shankar Questions About Quantum Mechanics

    Thanks for the explanation, but it's actually a bit too difficult.
    I never saw Hilbert spaces.

    I was just wondering what happens if you multipy |i> with <i|V>

    Do you get <i| i | V > ?

    That is not clear to me.

    @ jmcelve, I'll read your post later ( christmas dinner) anyway still thanks for the help.
     
  6. Dec 24, 2012 #5

    BruceW

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    Re: Shankar Questions About Quantum Mechanics

    merry christmas, hohoho

    <i|V> is just a number (Since you are taking an inner product). So then to multiply by <i| just means that you have <i| times by a number
     
  7. Dec 25, 2012 #6
    Re: Shankar Questions About Quantum Mechanics

    it is just like writing a vector in terms of it's basis.since the basis are in general infinite in number,it is called hillbert space.
     
  8. Dec 28, 2012 #7
    Re: Shankar Questions About Quantum Mechanics

    Thx for the help guys.

    However now I'm stuck with this proof:

    SenOx.jpg


    Aplpy axion 1(i), they say, but there is no such axion labelled that way, so I don't even know what axion to apply.

    If you can explain 1.3.18 a bit more, because I completely do not understand it, it would be great.

    Sorry for the questions ( it's no homework, it's self-study)

    Thx in advance.
     
  9. Dec 28, 2012 #8

    Doc Al

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    Re: Shankar Questions About Quantum Mechanics

    While not labeled as such, I assume they mean the axioms bullet-listed on page 8 (in my copy). The first one is skew-symmetry, which is what is invoked here.
     
  10. Dec 28, 2012 #9
    Re: Shankar Questions About Quantum Mechanics

    But why are there 3 terms in the first step and 4 terms in the second step @1.3.18

    pOsyZ.jpg

    Why is this not legit?
     
  11. Dec 28, 2012 #10

    Doc Al

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    Re: Shankar Questions About Quantum Mechanics

    What 3 terms?
     
  12. Dec 28, 2012 #11
    Re: Shankar Questions About Quantum Mechanics

    next to <Z|Z> ?
     
  13. Dec 28, 2012 #12

    Doc Al

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    Re: Shankar Questions About Quantum Mechanics

    I see 4 terms, not 3.
     
  14. Dec 28, 2012 #13
    Re: Shankar Questions About Quantum Mechanics

    I really don't understand 1.3.18 Doc Al.

    Proof in Griffiths is totally different and that, I understand, but this one in Shankar, not a single clue :(.

    I dont even understand how you get the <Z|Z> out of 1.3.17, sigh.
     
  15. Dec 28, 2012 #14
    Proof Schwarz-Inequality help

    Hi there,

    I'm reading Shankar, but I'm really stuck on this one:

    http://i.imgur.com/SenOx.jpg

    The whole equations @ 1.3.18 are complete gibbrish to me.

    Maybe somebody can explain this?
     
  16. Dec 28, 2012 #15

    mfb

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    Re: Proof Schwarz-Inequality help

    In the first step, 1.3.17 is used to replace Z, and the second step uses linearity (more precise: sesquilinearity) of the scalar product. With the help of 1.3.19, you can see that this has to be real and at least 0.
     
  17. Dec 28, 2012 #16
    Re: Proof Schwarz-Inequality help

    And why is this wrong:

    pOsyZ.jpg

    What is <Z| ?
     
  18. Dec 28, 2012 #17

    Doc Al

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    Re: Shankar Questions About Quantum Mechanics

    The first line just expresses <Z|Z> in terms of the what Z is defined as in 1.3.17. Then just multiply it out. You get four terms (not three--I don't know where you got three from). Do you understand where each of the four terms come from?
     
  19. Dec 28, 2012 #18
    Re: Proof Schwarz-Inequality help

    The second step I got (just worked it out), I know get u have to complex conjugate it.

    the first step I dont get, how u get <Z|
     
  20. Dec 28, 2012 #19
    Re: Shankar Questions About Quantum Mechanics

    Now I understand where the 4 vectors come from. It was the formula of antilinearity :)

    But I dont know how u get <Z|Z>

    I think <Z| = http://i.imgur.com/pOsyZ.jpg ( second line)

    I know i'm doing something impossible here, but not what.
    Working with brakets first time ever, is very very confusing.


    In other words: I think you have got to reverse the left Z in <Z|Z> , and tthus also the equation in 1.3.17 , because its a bra, while the right Z is a ket.
     
  21. Dec 28, 2012 #20

    Doc Al

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    Re: Shankar Questions About Quantum Mechanics

    Moderator's note: I merged the two threads. (Once is enough!)​
     
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