1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expectation value of an operator (not its corresponding observable value)

  1. Feb 11, 2007 #1
    1. Problem statement
    This isn't a homework question itself, but is related to one. More specifically, I'm computing the time-derivative of [tex]\langle x \rangle[/tex] using the correspondence principle. One side simplifies to [tex]\left\langle \frac{\hat{p}}{m} \right\rangle[/tex], but what is the physical meaning of this? How does one compute the expectation value of an operator? The concept is alien to me.

    2. Relevant equations
    [tex]\langle Q \rangle = \int_{-\infty}^{\infty}\Psi^* \hat{Q} \Psi \; dx[/tex]
    [tex]\langle \hat{Q} \rangle = ???[/tex]
     
  2. jcsd
  3. Feb 12, 2007 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The expectation value of an operator A in a pure quantum state [itex] |\psi\rangle [/itex] is the complex number

    [tex] \langle A\rangle_{|\psi\rangle} =\langle \psi|A|\psi\rangle [/tex]
     
  4. Feb 12, 2007 #3
    Well yes, but that's simply the generalized form of the first equation I posted. So what you're saying is that they are the same? [tex]\left\langle \frac{\hat{p}}{m} \right\rangle = m^{-1}\int_{-\infty}^{\infty}\Psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi \; dx[/tex]?
     
  5. Feb 12, 2007 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    That is correct. The generalization doesn't single out the (rigged) Hilbert space. That's the advantage of using abstract bra-ket notation.
     
  6. Feb 12, 2007 #5
    That makes sense. I mistakenly thought there is a difference between [tex]\langle Q \rangle[/tex] and [tex]\langle \hat{Q} \rangle[/tex], but that may have resulted from an abuse of notation.

    Thanks a lot, dextercioby.
     
  7. Oct 4, 2011 #6
    Actually, what is difference between <Q> and <Q_hat> ?
     
  8. Oct 4, 2011 #7

    G01

    User Avatar
    Homework Helper
    Gold Member

    Read the three posts above your post, again. The OP and dextercioby came to the conclusion that there was no difference between the two. The OP's confusion resulted from an abuse or misuse of the notation involved.

    Also, this thread is almost 5 years old! If you still have questions on this topic, please start a new thread.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Expectation value of an operator (not its corresponding observable value)
Loading...