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Product rule of derivative of expectation values

  1. Jun 11, 2013 #1
    Hello, first post here.

    I am preparing for my Introductory Quantum Mechanics course, and in the exam questions, we are asked to use Ehrenfest's theorem to show that
    [tex] \frac{d}{dt}\langle \vec{r}\cdot \vec{p} \rangle = \langle 2T-\vec{r}\cdot \nabla V \rangle [/tex]

    Now, from other results:
    [tex] \frac{d}{dt}\langle \vec{r} \rangle = \frac{1}{m} \langle \vec{p} \rangle [/tex]
    [tex] \frac{d}{dt}\langle \vec{p} \rangle = \langle -\nabla V \rangle [/tex]

    it can be solved using a product rule of differentiation:
    [tex] \frac{d}{dt}\langle \vec{r}\cdot \vec{p} \rangle =\frac{d}{dt}\langle \vec{r} \rangle \cdot \vec{p} + \vec{r} \cdot \frac{d}{dt}\langle \vec{p} \rangle [/tex]
    But I severely doubt this is the right way, or even mathematically sound. So my question is, is the above equation sound, or should I be calculating the commutators with the Hamiltonian, as per Ehrenfest's?
     
    Last edited: Jun 11, 2013
  2. jcsd
  3. Jun 11, 2013 #2

    Bill_K

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    That equation's fine, as long as you keep the operators in their original order, which you did.
     
  4. Jun 11, 2013 #3
    Thanks. Google failed me, so I couldn't find any documentation. Happy to know my instincts weren't all off.
     
  5. Jun 12, 2013 #4

    Jano L.

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    Gold Member

    I think the formula that follows the above words is not correct. On the left-hand side, there is a number, while on the right-hand side, there is an operator (##\mathbf{p} = -i\hbar \nabla##).

    Your intuition is right, you can derive the virial theorem by using the time-dependent Schroedinger equation and the commutation relations. Here is how:


    $$
    \frac{d}{dt}\langle \mathbf r\cdot\mathbf p\rangle = \int \dot \psi^* \mathbf r\cdot\mathbf p \psi + \psi^* \mathbf r\cdot\mathbf p \dot \psi\,dV
    $$

    Now use Schroedinger's equation

    $$
    \dot \psi = \frac{1}{i\hbar} {H} \psi
    $$

    to get rid of time derivatives. The integrand contains the commutator

    $$
    [\mathbf r\cdot\mathbf p, H].
    $$

    Replace ##H## by ##T+V## and use the commutation relations for position and momentum operator and the relation
    $$
    [A, BC] = B[A,C] + [A,B]C.
    $$
    After some manipulations, you should get ##\langle 2T - \mathbf r\cdot \nabla V\rangle##.
     
  6. Jun 12, 2013 #5

    vanhees71

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    2016 Award

    You cannot use the product rule so easily, because the expectation value of a product is not the product of the expectation value. However, Ehrenfest's theorem tells you that
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle=\left \langle \frac{\mathrm{D} \hat{A}}{\mathrm{D} t} \right \rangle.[/tex]
    For a not explicitly time-dependent operator the covariant time derivative is given by
    [tex]\frac{\mathrm{D} \hat{A}}{\mathrm{D} t}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}].[/tex]
    Since, mathematically the commutator is a derivation, here the product rule holds.
    [tex]\frac{\mathrm{D} \hat{\vec{r}} \cdot \hat{\vec{p}}}{\mathrm{D} t} = \frac{\mathrm{D}\hat{\vec{r}}}{\mathrm{D} t} \cdot \hat{\vec{p}} + \hat{\vec{r}} \cdot \frac{\mathrm{D}\hat{\vec{p}}}{\mathrm{D} t}.[/tex]
    Now, for a usual non-relativistic Hamiltonian,
    [tex]\hat{H} = \frac{\hat{\vec{p}}^2}{2m} + V(\hat{\vec{r}}),[/tex]
    you get
    [tex]\frac{\mathrm{D} \hat{\vec{r}}}{\mathrm{D} t}=\frac{\hat{\vec{p}}}{m}, \quad \frac{\mathrm{D} \hat{\vec{p}}}{\mathrm{D} t} = -\vec{\nabla} V(\hat{\vec{r}}).[/tex]
    Using
    [tex]\hat{T}=\frac{\hat{\vec{p}}^2}{2m}[/tex]
    for the kinetic energy, this gives the virial theorem.

    Note, however, that this makes only sense as a property of physical observables, when you apply this to the self-adjoint operator,
    [tex]\frac{1}{2}(\hat{\vec{r}} \cdot \hat{\vec{p}}+\hat{\vec{p}} \cdot \hat{\vec{r}}).[/tex]
     
  7. Jun 13, 2013 #6

    dextercioby

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    The problem is mathematically ill-posed, since Ehrenfest's theorem is formulated in terms of self-adjoint operators, while the operator [itex] \mathbb{r}\cdot\mathbb{p} [/itex] is not even symmetric, hence, if the equality is correct, it can't be proven using Ehrenfest's theorem.
     
    Last edited: Jun 13, 2013
  8. Jun 14, 2013 #7
    Thanks for the answers. I ended up using Ehrenfest's theorem, and straight-forward calculation of the commutator with the Hamiltonian.
    @dextercioby: So you're saying Ehrenfest doesn't even apply here? Is it only true in general for self-adjoint operators? Do you have an example of a case where there is a clear discrepancy between the answer from Ehrenfest and the correct answer?

    EDIT: I see now vanhees mentioned the same at the end of his post. Is this a standard method for making non-hermitian operators self-adjoint?
     
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