Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]
    [itex]<\psi\frac{d}{dq}=-<\frac{d\psi}{dq}[/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

    Bill_K

    User Avatar
    Science Advisor

    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
    [itex]\overline{\frac{d}{dq}\psi>}=\overline{\frac{d\psi}{dq}>}=<\frac{d\overline{\psi}}{dq}[/itex]

    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>}=<\psi\overline{\frac{d}{dq}}[/itex]
    [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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Few questions on Dirac's Principles of QM.
  1. A few QM questions. (Replies: 11)

  2. A few QM questions (Replies: 2)

Loading...