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Few questions on Dirac's Principles of QM.

  1. Sep 8, 2011 #1
    Hi there,

    I'm reading Dirac's Principles of QM, but I think I miss something...

    1)When he derives the Poisson quantic bracket, he states that [itex]u_1v_1-v_1u_1=i\hbar[u_1,v_1][/itex] and says that hbar must be real since we introduced the imaginary unit. The thing is, since uv-vu is real, because we are talking about dynamics variable that are real linear operator, this means that [u,v] must be pure imaginary. But that doesn't fit well with what Dirac says a few lines below, where he hypothesizes that [u,v] in QM has the same value than in classical mechanic (and I derive from here that it is real).
    What am I missing?

    2)Why is so important a representation where the observables are diagonal? And why it is so important to build Schroedinger's representation?

    3)I can't follow his derivation of the conjugate transpose of d/dq :
    giving that [itex]\frac{d}{dq}\left.\psi\right>=\left.\frac{d\psi}{dq}\right>[/itex]
    he makes these consideration: the conjugate transpose of [tex]\frac{d}{dt}\psi>\,\,\mathrm{is}\,\,<\frac{d\bar{\psi}}{dq}[/tex] and, from the previous equation, [itex]<\frac{d\bar{\psi}}{dq}=-<\bar{\psi}\frac{d}{dq}[/itex] hence the conjugate transpose of d/dq -d/dq.
    My question is:why [itex]\bar{\frac{d\psi}{dq}>}=<\frac{d\bar{\psi}}{dq}[/itex]? I thought that [itex]\forall |P>, \bar{|P>}=<P|[/itex]and this gives [itex]\bar{\frac{d\psi}{dq}>}=<\frac{d\psi}{dq}[/itex]
    Again, what am I missing?

    4) What does it mean that [itex]\mathbf{p}=i\hbar\mathbf{d}[/itex], where d is the translation operator? Better, what is the physical importance and meaning?

    Thank you in advance, more questions coming :D
  2. jcsd
  3. Sep 8, 2011 #2


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    I think much of your question is a result of terminology. Where you say a physical quantity needs to be represented by a "real linear operator", what you mean is Hermitian or self-adjoint. The Hermitian conjugate involves a complex conjugate and an order reversal, probably what you're calling conjugate transpose. Note that if u and v are Hermitian, then uv - vu will be anti-Hermitian. This explains the appearance of i in question 1.

    It also explains question 3, since ∂/∂q is anti-Hermitian also. When you take the adjoint of ∂/∂q |ψ> you'll pick up a minus sign.
  4. Sep 8, 2011 #3
    Ok, I get (probably) the answer to question (1), but not to the others.

    Just to clarify: is the equation [itex]\overline{\alpha|P>}=<P|\overline{\alpha}[/itex](*) correct, the bar meaning taking the complex conjugate and alpha being an hermitian operator?
    If so, then I can't understand why the following passage is right

    If we consider in (*) |P>=dP/dq>, then it should be [itex]\overline{\frac{d\psi}{dq}>}=<\frac{d\psi}{dq}[/itex] w/o the complex conjugate on psi.

    Can it be this way:
    [itex]\overline{\frac{d}{dq}\psi>}=\overline{\frac{d\psi}{dq}>}=<\frac{d\psi}{dq}=-<\psi\frac{d}{dq}[/itex] hence [itex]\overline{\frac{d}{dq}}=-\frac{d}{dq}[/itex]? (This demonstration is different from Dirac's but arrives to the same conclusion).
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