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Expectation values of Kin energy in Bra-Ket notation

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Confirm explicitly that ##\frac{1}{2m}\langle \hat{p}_x \Psi | \hat{p}_x \Psi \rangle## cannot be negative.

    2. Relevant equations

    ##-i\hbar \frac{\partial}{\partial x} = \hat{p}_x##

    3. The attempt at a solution

    i seem to get:
    ##\frac{1}{2m}\langle \hat{p}_x \Psi | \hat{p}_x \Psi \rangle=\frac{\hbar^2}{2m}\left ( \frac{\partial \Psi}{\partial x}\frac{\partial \Psi}{\partial x}\right )##
    This seems ok to me...am I right in thinking that because one of the momentum operators is in the BRA, then it has to be quantum conjugated and becomes ##i\hbar\frac{\partial}{\partial x} = \hat{p}_x^*##?
     
  2. jcsd
  3. May 19, 2013 #2

    Fredrik

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    The statement of this problem confuses me. First of all, that's not bra-ket notation. (Bra-ket notation has the operator in the center, like this: <x|A|y>). It looks like an inner product to me. And it follows immediately from the definition of "inner product" that this is non-negative (assuming that m is non-negative). But I guess you're not supposed to use that, since that would make the problem too trivial.

    So we have to figure out what "explicitly" means here. I would guess that it means that you're supposed to use the definition of this particular inner product.
     
  4. May 19, 2013 #3
    Thanks for your help with this. In the end I've inserted a general wavefunction and demonstrated that the complex parts reduce to 1 and the real parts (derivative of the wavefunction) are effectively squared, thereby not allowing negative results. I'm not sure if this is what they are after.

    I'm interested by your question regarding this being bra-ket notation. This expression started off as an integral over all space of ##\Psi^* \frac{\hat{P}_x}{2m} \Psi## which I adapted to bra-ket notation as $$\Psi^* \frac{\hat{P}_x}{2m} \Psi=\frac{1}{2m}\langle \Psi | \hat{P}_x \cdot \hat{P}_x | \Psi \rangle$$

    I then moved the momentum operator to the bra vector because it is a linear hermitian operator. Is this not correct?
     
  5. May 19, 2013 #4

    Fredrik

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    When we're using bra-ket notation, the only way we can "move the operator to the bra" is to use associativity:
    $$\langle f|\left(A|g\rangle\right) =\left(\langle f|A\right)|g\rangle.$$ Note that here |g> is a vector, A is a linear operator, and <f| is a linear functional that takes vectors (i.e. kets) to complex numbers. The right-hand side above is defined by saying that it's equal to the left-hand side.

    If I use the notation (x,y) for the inner product of two vectors x and y, the left-hand side of the equality above by definition means
    $$\left(\left|f\rangle,A|g\rangle\right.\right).$$ By definition of ##A^\dagger##, this is equal to $$\left(A^\dagger\left|f\rangle,|g\rangle\right.\right).$$ If we define |h> by ##|h\rangle=A^\dagger |f\rangle##, we can rewrite this in bra-ket notation as
    $$\langle h|g\rangle.$$ Since these equalities hold for all |g>, they imply that ##\langle f|A=\langle h|##. So now we know that the bra that corresponds to the ket ##A^\dagger|f\rangle## is ##\langle f|A##, and because of that, we can interpret the first equality I wrote down as saying that
    $$\left(\left|f\rangle,A|g\rangle\right.\right) =\left(A^\dagger\left|f\rangle,|g\rangle\right.\right).$$
     
    Last edited: May 19, 2013
  6. May 19, 2013 #5
    Hmm, well in that case I have got the first part wrong.

    The first part of the question asks us to use the fact that ##\langle A^2 \rangle = \langle \hat{A} \Psi | \hat{A} \Psi \rangle## where the operator is linear and Hermitian and the wavefunction represents a bound state to show that:
    ##\langle E_{kin} \rangle = \frac{1}{2m} \langle \hat{P} \Psi | \hat{P} \Psi \rangle##

    I just started from ##\hat{H}=\frac{\hat{P}^2}{2m}##, argued that the square of the momentum operator was the product of two Hermitian operators (and therefore Hermitian itself) and used that reasoning to move one of the momentum operators to the bra side, complex conjugating the constant coefficient. It gives the right answer, but is obviously the wrong reasoning.

    So do you think I actually need to show that the operator ##\hat{P}^2## can be expressed as an inner product? I'm not really sure how to do that...
    Thanks again for your time.

    EDIT: I've just checked my textbook, and it states that:
    "If any operator ##\hat{A}## satisfies the condition ##\langle f | \hat{A}g \rangle = \langle \hat{A} f | g \rangle## for any normalizable functions f(x) and g(x), then the operator is Hermitian."

    It also says that any observable quantity in QM is represented by a linear Hermitian operator.
    Surely those two statements are enough to equate ##\langle \Psi | \hat{P}\cdot\hat{P} | \Psi \rangle## to ##\langle \hat{P} \Psi | \hat{P} \Psi \rangle##?
     
    Last edited: May 19, 2013
  7. May 19, 2013 #6

    Fredrik

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    You're not doing anything wrong except calling the notation "bra-ket notation". You're working with a notation for inner products of vectors, not a notation for bras and kets.

    What book are you using? I'm curious about what it says about bra-ket notation.
     
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