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Commutation between operators of different Hilbert spaces

  1. Aug 11, 2015 #1
    Hi!

    If I have understood things correctly, in a multi-electron atom you have that the spin operator ##S## commutes with the orbital angular momentum operator ##L##. However, as these operators act on wavefunctions living in different Hilbert spaces, how is it possible to even calculate the commutator?

    I mean, you can't calculate the commutator between ##\hat{H} = i \hbar \frac{\partial}{\partial t}## and ##\hat{p}= \frac{\hbar}{i} \vec{\nabla}## precisely because of different hiblert spaces, so why is ##L## and ##S## different?
     
  2. jcsd
  3. Aug 11, 2015 #2

    ShayanJ

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    Gold Member

    At first note that usually ##\hat H## is used for the Hamiltonian which isn't identical with the temporal derivative operator.
    Anyway, what you say here is wrong You surely can calculate that commutator and its really straight-forward. You just should consider the position-space wave-function together with its temporal dependence and calculate ## [ i \hbar \frac{\partial}{\partial t},\frac{\hbar}{i} \vec{\nabla}]\Psi(\vec x,t)##. Also there is only one Hilbert space here, the space of square-integrable complex valued functions.

    And about the commutator of spin and orbital angular momentum operators. Here there are two Hilbert spaces and the abstract quantum state of the system is a member of the tensor product of these two Hilbert spaces, so we have ## |\Psi\rangle=\sum_{nm} c_{nm}|\Phi_n\rangle\otimes |\Sigma_m\rangle## where ## \{|\Phi_n\rangle\}## is a basis for the spatial Hilbert space and ## \{|\Sigma_m\rangle\}## is a basis for the spin Hilbert space. The operators are tensor products of operators acting on each Hilbert space too. So instead of ## \hat L ## or ## \hat S##, we should have ## \hat L \otimes \hat I_{spin} ## and ## \hat I_{spatial} \otimes \hat S##.
    Now we can calculate the commutator: ## [\hat L \otimes \hat I_{spin},\hat I_{spatial} \otimes \hat S]=[\hat L,\hat I_{spatial}]\otimes [\hat I_{spin},\hat S]##.
     
  4. Aug 12, 2015 #3

    aleazk

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    Gold Member

    What @Shyan says is correct. But I think that, for practical calculations, the following point of view is more convenient.

    The tensor product Hilbert space ##L^{2}(\mathbb{R}^{3})\otimes\mathbb{C}^{2}## of the two spaces ##L^{2}(\mathbb{R}^{3})## and ##\mathbb{C}^{2}## is isomorphic to the Hilbert space ##L^{2}(\mathbb{R}^{3},\mathbb{C}^{2})##, which is the space of square integrable vector-valued (over ##\mathbb{C}^{2}##; in the integral expression of the inner product, one uses the natural inner product on ##\mathbb{C}^{2}## rather than the complex modulus of ##\mathbb{C}##) functions on ##\mathbb{R}^{3}##.

    In this way, both the orbital angular momentum and the spin operators act always on these vector-valued functions on ##\mathbb{R}^{3}##. But the angular momentum operators are given by the usual differential operators and act componentwise, while the spin operators are given by the usual matrices and act on the components of the functions (and thus mixing them), i.e.,

    $$
    S\left(\begin{array}{c}
    A\\
    B
    \end{array}\right)=\left(\begin{array}{cc}
    a & b\\
    c & d
    \end{array}\right)\left(\begin{array}{c}
    A\\
    B
    \end{array}\right)=\left(\begin{array}{c}
    aA+bB\\
    cA+dB
    \end{array}\right)$$ and $$
    L\left(\begin{array}{c}
    A\\
    B
    \end{array}\right)=\left(\begin{array}{c}
    LA\\
    LB
    \end{array}\right)$$

    You can easily check that the two operators commute.
     
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