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How to understand current-density operator

  1. Mar 28, 2013 #1
    Dear all,
    There is a definition of current-density operator, and the form is as follows:
    I cannot understand this form, because i think ∇ operator and δ operator are commutable.
    Another form of current-density operator can be found from this website:
    Are these two forms equal?
    Last edited: Mar 28, 2013
  2. jcsd
  3. Mar 28, 2013 #2
    The gradient with respect to some variable, and any function of that variable, are not going to commute in general, because in one case you multiply the wavefunction by the function-operator and take grad of the result, and in the other case you take grad of the wavefunction and multiply by the function-operator afterwards. This is, for example, how the real-space co-ordinate definition of the momentum operator can be shown to reproduce the [x,p] canonical commutation relation. Hope that helps.
  4. Mar 28, 2013 #3


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    The second definition which you took from wikipedia is not for the operator but for it's expectation value, at least in first quantization, i.e. as long as psi is not considered to be an operator itself. As psmt has pointed out already, the grandient and the delta function don't commute with each other, or more specifically ##[\nabla \delta(x-x_0)\delta(y-y_0)\delta(z-z_0)]=(\delta'(x-x_0)\delta(y-y_0)\delta(z-z_0),\delta(x-x_0)\delta'(y-y_0)\delta(z-z_0),\delta(x-x_0)\delta(y-y_0)\delta'(z-z_0))^T##, where ##\delta'(x)=d/dx \delta(x)##.
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