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Relation between commutation and quantization

  1. Mar 31, 2012 #1
    relation between "commutation" and "quantization"

    Hi people;

    Over the several texts I have read, I got the impression that position-momentum commutation relations is the cause of "quantization" of the system. Or, they are somehow fundamentally related.

    The only relation I know of, is to derive the momentum operator in position space, [itex]-i\hbar\frac{d}{dx}[/itex], from the commutation relation [itex][x,p]=i\hbar[/itex], and then find the position and momentum eigenfunctions which turn out to be oscillating functions of [itex]x[/itex] and [itex]p[/itex]. Then, eigenvalue spectrum of these operators are then naturally "quantized", BUT only if the potential is bounding, like box, harmonic oscillator etc..

    Now this demonstration of relation between commutation and quantization looks quite "indirect" to me, and also it is conditional (a bounding potential required to get quantized eigenvalues).

    So my question is; is there a more fundamental demonstration of the relation between commutation relations and quantization of a system.

    Thanks in advance for the answers.

  2. jcsd
  3. Mar 31, 2012 #2


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    Re: relation between "commutation" and "quantization"

    But, the commutation relation still holds between x and p when x and p are not quantized, for a free particle wave packet.
  4. Apr 1, 2012 #3
    Re: relation between "commutation" and "quantization"

    Exactly.. I want to know if the notion that I've got, that commutation and quantization is fundamentally related to each other, is true or not.

    If it is true; then, is there a mathematical way to show that in a more general way than the example that I gave above.
  5. Apr 1, 2012 #4


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    Re: relation between "commutation" and "quantization"

    Quantization means replacing classical functions on phase space, here x and p by, QM operators; in position space p becomes an operator -id/dx.

    The commutation relation for the operators x and p can derived from this position space representation, i.e.

    [x, -id/dx] f(x) = (id/dx x) f(x) = i f(x) for all f(x), so [x, -id/dx] = i
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