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B Position operator

  1. Sep 2, 2016 #1
    Well i am noobie to quantum physics so i matbe totally incorrect so please bear with me.
    I had question how is position operator defined mathematically.
    I was reading the momentum position commutator from http://ocw.mit.edu/courses/physics/...pring-2013/lecture-notes/MIT8_04S13_Lec05.pdf (page 2 of pdf)
    They have used position operator = eigenvalue (ie position itself) times wavefunction http://ocw.mit.edu/courses/physics/...pring-2013/lecture-notes/MIT8_04S13_Lec05.pdf

    But i doubt that the relation will be valid only for delta wavefunction (paralell to as momentum relation is valid in case of eix. I understood it as momentum is well defined in only that case so similarly position will be defined clearly in delta function only.)
    So am i correct, also point pitfalls in my understandings
     
  2. jcsd
  3. Sep 2, 2016 #2

    vanhees71

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    In the wave-mechanics formulation (the position representation of Hilbert space) you associate with the Hilbert space vector a function ##\psi(\vec{x})##, which is square integrable, i.e.,
    $$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(\vec{x})|^2$$
    exists. Such functions build a Hilbert space (let's leave out the mathematicle subtlties here), the space of square integrable functions. The scalar product is defined by
    $$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi_1^*(\vec{x}) \psi_2(\vec{x}),$$
    which always exists for ##\psi_1## and ##\psi_2## being square integrable.

    Now you can easily check that the operators ##\hat{x}_i## and ##\hat{p}_i##, defined by
    $$\hat{x}_i \psi(\vec{x})=x_i \psi(\vec{x}), \quad \hat{p}_i \psi(\vec{x})=\frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x_i} \psi(\vec{x})$$
    obey the commutator relations for position and momentum,
    $$[\hat{x}_i,\hat{x}_j]=0, \quad [\hat{p}_i,\hat{p}_j]=0, \quad [\hat{x}_i,\hat{p}_j]=\mathrm{i} \hbar.$$

    Now the eigenvalue problem for such operators is a bit more complicated than for operators in a finite-dimensional vector space. Take the momentum operator as an example. The eigenvalue equation reads
    $$\hat{p}_j u_{\vec{p}}(\vec{x})=p_j u_{\vec{p}}(\vec{x}).$$
    You can solve this equation, using the definition of the momentum operator easily to be
    $$u_{\vec{p}}(\vec{x})=N \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right), \quad N=\text{const}.$$
    But now you see that for any ##\vec{p} \in \mathbb{R}^3## this is not a square-integrable function, i.e., it's not in the Hilbert space! Rather it's a distribution (or generalized function). You can formaly evaluate a scalar product of two such generalized eigenfunctions to give
    $$\langle u_{\vec{p}}|u_{\vec{p}'} \rangle = |N|^2 (2 \pi)^3 \hbar^3\delta^{(3)}(\vec{p}-\vec{p}').$$
    So it's convenient to define
    $$N=\frac{1}{(2 \pi \hbar)^{3/2} }.$$
    A similar argument leads to the "position eigenvectors". The position eigenvector to eigenvalue ##\vec{x}_0## must be
    $$u_{\vec{x}_0}(\vec{x})=\delta^{(3)}(\vec{x}-\vec{x}_0).$$
    So again it's a distribution. You cannot even square it!

    Nevertheless you can use the generalized eigenvectors for very important calculations. E.g., if you have given a particle to be in a state represented by the (square integrable!) wave function ##\psi## and want to know the probaility distribution for momentum, you just evaluate formally
    $$\tilde{\psi}(\vec{p})=\langle u_{\vec{p}}|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}}^*(\vec{x}) \psi(\vec{x}),$$
    which just is the Fourier transform. If ##\psi## is normalized to 1, you get the momentum-probability distribution via Born's rule as
    $$P(\vec{p})=|\tilde{\psi}(\vec{p})|^2.$$
     
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