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Zero-zero transitions

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that the matrix element of [itex]\mathbf{p} \cdot \mathbf{A} [/itex] between 1S and 2S vanishes to all orders.
    I think I need to show the the following

    [itex]\langle 2\,0\,0 \mid \boldsymbol{\epsilon}^*(\mathbf{k}, \lambda) \cdot \mathbf{p} e^{-i\mathbf{k}\cdot \mathbf{r}} \mid 1\,0\,0 \rangle = 0[/itex]


    2. Relevant equations
    By the orthogonality of [itex]\boldsymbol{\epsilon}^*[/itex] and [itex]\mathbf{k}[/itex] we can change the order
    [itex]\left[ \boldsymbol{\epsilon}^*(\mathbf{k}, \lambda) \cdot \mathbf{p},\, e^{-i\mathbf{k}\cdot \mathbf{r}} \right] = 0[/itex]


    3. The attempt at a solution
    I had thoughts along parity lines, but I haven't been successful so far. Since the states are spherically symmetric, I can choose [itex]\mathbf{k}[/itex] to be along the z axis, which might lead to some clarity. Then we could have [itex]\boldsymbol{\epsilon}^*(k\hat{\mathbf{z}}, \pm) = \frac{1}{\sqrt{2}}\left(\hat{x} \mp i \hat{y} \right)[/itex].

    [itex]\langle 2\,0\,0 \mid \left(p_x \mp ip_y\right) e^{-ikz} \mid 1\,0\,0 \rangle[/itex]
    [itex]\langle 2\,0\,0 \mid \left(p_x \mp ip_y\right) \left(\cos(kz) - i \sin(kz)\right) \mid 1\,0\,0 \rangle[/itex]
    [itex]\langle 2\,0\,0 \mid p_x\cos(kz) \pm p_y\sin(kz) \mid 1\,0\,0 \rangle + i\dots[/itex]

    So it seems then I must have independently

    [itex]\langle 2\,0\,0 \mid p_x\cos(kz)\mid 1\,0\,0 \rangle =0 [/itex] and [itex]\langle 2\,0\,0 \mid p_y\sin(kz)\mid 1\,0\,0 \rangle =0 [/itex]

    I'm really not sure why this would be.
     
    Last edited: Apr 9, 2014
  2. jcsd
  3. Apr 9, 2014 #2
    You are not going into right direction.If you want to show it using the reasoning of parity change and so,then this will work only in the case of first order approximation i.e. dipole approximation.But you can see that there are infinite terms to all orders,arising from taylor expansion of exponential.And you have to check magnetic dipole transition and all other higher orders(which can be explained by angular momentum conservation).This is not what you should do,generally.You need to write the matrix element as a 3-dimensional integral and use explicitly the 1s and 2s state wavefunction of hydrogen.Then you can show the vanishing of integral as vanishing of matrix element between the two state.(actually you can show it without using the explicit form of wavefunctions but you have to see the symmetry of the integral).Also p is not acting on any eigenstate,so you should write the operator form of p acting on 1s state in matrix element evaluation i.e. p=-ih-
     
  4. Apr 9, 2014 #3
    Actually, I wasn't going very far in the wrong direction. For [itex]\langle 2\,0\,0 \mid p_x\cos(kz)\mid 1\,0\,0 \rangle[/itex], just integrate along the x direction first and get zero. Similarly for [itex]\langle 2\,0\,0 \mid p_y\sin(kz)\mid 1\,0\,0 \rangle [/itex], just integrate along y. In both cases [itex]p_i \mid 1\,0\,0 \rangle[/itex] is anti-symmetric about the [itex]i[/itex] direction. I think my difficulty was thinking of overall parity, instead of just along one direction.
     
    Last edited: Apr 9, 2014
  5. Apr 9, 2014 #4
    those p's are operators and will act on the states,you need to write it in operator form and matrix element will be simultaneously evaluated.Those operator acting on state will give give some dependence on r with which you will take the dot product with polarization vector.You are taking the product with operator itself without knowing where to act with it.
     
  6. Apr 10, 2014 #5
    I don't know what you mean.
     
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