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Dirac Notation and commutation

  1. Nov 6, 2013 #1
    Hi everyone, my problem is this
    Using Dirac notation show that

    [itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|[itex]\hat{A}[/itex]|[itex]\varphi[/itex]> = [itex]\frac{i}{\hbar}[/itex]<[itex]\varphi[/itex]|[[itex]\hat{H}[/itex],[itex]\hat{A}[/itex]]|[itex]\varphi[/itex]>

    where A does not explicitly depend on t




    I am given as a hint that the hamiltonian operator in Dirac notation is
    i[itex]\hbar[/itex][itex]\frac{d}{dt}[/itex]|[itex]\varphi[/itex]>=[itex]\hat{H}[/itex]|[itex]\varphi[/itex]>

    and that the conjugate expression is

    -i[itex]\hbar[/itex][itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|=<[itex]\varphi[/itex]|[itex]\hat{H}[/itex]




    I'm a bit stumped, I know that the left hand side of the equation is the time derivative of the expectation value of operator A but as for the right hand side. I tried to expand the commutator as normal as if it were acting on the function (ket) |phi> but this came out to be zero which if true would surely make the whole RHS zero, which makes me think I don't understand what the rhs means. Any help would be appreciated.
     
  2. jcsd
  3. Nov 6, 2013 #2

    CAF123

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    Gold Member

    Perhaps start with the LHS first and rewrite $$\frac{d}{dt} \langle \hat{A} \rangle = \frac{d}{dt} \langle \psi | \hat{A} | \psi \rangle = \frac{d}{dt} \int dx\, \psi^* \hat{A}\, \psi$$

    The RHS can be zero but only if ##\hat{A}## and ##\hat{H}## were to commute, but this was not given.
     
    Last edited: Nov 6, 2013
  4. Nov 6, 2013 #3
    Use the product rule one the left hand side first.
    Then impose the hint to rewrite some of the terms.
    It should become quite clear almost immediately.

    An extra question, does it change a lot if the operator A does explicitly depend on time?
     
  5. Nov 6, 2013 #4
    Thanks to both of you, I think I have it

    [itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|[itex]\hat{A}[/itex]|[itex]\varphi[/itex]> = <[itex]\varphi[/itex]|[itex]\hat{A}[/itex][itex]\frac{d}{dt}[/itex]|[itex]\varphi[/itex]> + [itex]\hat{A}[/itex][itex]\varphi[/itex]>[itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|

    then from the hints

    [itex]\frac{d}{dt}[/itex](|[itex]\varphi[/itex]>) = [itex]\frac{\hat{H}}{i\hbar}[/itex]( |[itex]\varphi[/itex]>) and [itex]\frac{d}{dt}[/itex](<[itex]\varphi[/itex]|) = -[itex]\frac{\hat{H}}{i\hbar}[/itex]( <[itex]\varphi[/itex]|)

    Substitute those in, do a bit of simplifying and rearranging then I got the correct answer. Thanks a bunch

    I do have another question if I may

    I have to show how we get the commutation identity [[itex]\hat{A}[/itex],[itex]\hat{B^{2}}[/itex]]=[[itex]\hat{A}[/itex],[itex]\hat{B}[/itex]][itex]\hat{B}[/itex] + [itex]\hat{B}[/itex][[itex]\hat{A}[/itex],[itex]\hat{B}[/itex]] but am having trouble, I'd be very grateful for any suggestions in the right direction.

    Thanks again
     
  6. Nov 6, 2013 #5

    CAF123

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    Gold Member

    Try working with the RHS this time and show that it reduces to the LHS.
    Also, [itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|[itex]\hat{A}[/itex]|[itex]\varphi[/itex]> = <[itex]\varphi[/itex]|[itex]\hat{A}[/itex][itex]\frac{d}{dt}[/itex]|[itex]\varphi[/itex]> + [itex]\hat{A}[/itex][itex]\varphi[/itex]>[itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|

    in the second term, I think that should be ##\left( \frac{d}{dt} \langle \psi | \right) \, \hat{A} \psi,## when you are dealing with operators, the ordering matters.
     
    Last edited: Nov 6, 2013
  7. Nov 6, 2013 #6
    Thanks for the help with the last question, hadn't realised it would be so easy, however for my original problem, what I meant in that term you flagged up is

    ([itex]\hat{A}[/itex]|[itex]\varphi[/itex]>)([itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|)

    Is this not correct? I thought

    ([itex]\hat{A}[/itex]|[itex]\varphi[/itex]>)([itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|) = ([itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|)([itex]\hat{A}[/itex]|[itex]\varphi[/itex]>)

    Is that not right?

    Thanks again
     
  8. Nov 6, 2013 #7
    No it's not right. It's like [itex]\mathbf{v}^T\mathbf{w} \neq \mathbf{w} \mathbf{v}^T[/itex]

    The first results in a scalar (it's a representation of an inner product), the second results in a higher order tensor (or in other words it's an outer product).

    [itex]\langle a \mid b \rangle[/itex] is a scalar.
    [itex]\mid a \rangle \langle b\mid[/itex] is an operator that "eats" up [itex]\mid b \rangle[/itex] and results in some multiple of [itex]\mid a \rangle [/itex].
     
  9. Nov 7, 2013 #8
    First, if you assume the operator has an explicit time dependence, you can write:
    [itex] \frac{d}{dt}\langle \phi |A(x,t)|\phi\rangle = \langle \frac{d\phi}{dt}|A|\phi\rangle +\langle \phi|\frac{dA}{dt}|\phi \rangle +\langle \phi |A|\frac{d \phi}{dt}\rangle [/itex]

    Then, the R.H.S. can be simplified to:
    [itex] \frac{1}{-i\hbar}\langle \phi |HA|\phi \rangle + \frac{1}{i\hbar}\langle \phi |AH|\phi \rangle +\langle \frac{dA}{dt}\rangle[/itex]

    The final result is:
    [itex] \frac{i}{\hbar}\langle [H,A]\rangle + \langle \frac{dA}{dt}\rangle [/itex]

    Drop the dA/dt for no explicit time dependence.
     
    Last edited: Nov 7, 2013
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