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What is uncertainty relation

  1. Jul 24, 2014 #1

    One of the most asked questions is concerning how to derive the Heisenberg Uncertainty Relation.

    Starting from almost basic concepts of Quantum Mechanics, a derivation is given here. Some details are left as minor exercises for the interested reader.

    The derivation on this page is based on the integral version of the Schwarz inequality, applied to wavefunctions.

    An alternative derivation, based on the expectation value version, and applied to bras and kets, is on the page "uncertainty principle".


    [tex] \Delta x\Delta p \geq \hbar/2 [/tex]

    Extended explanation

    For understanding this derivation, one might need to pick up results from libary items on Schwarz inequality and hermitian operators.

    The commutator of two operators [itex]A [/itex] and [itex]B [/itex] is defined as:
    [tex] \left[ A,B \right] = AB - BA. [/tex]

    Some commutator algebra:
    [tex] \left[ A,B + C \right] = \left[ A,B \right] + \left[ A,C \right] . [/tex]
    [tex]\left[ A,q \right] = 0 ,[/tex] if [itex] q [/itex] is a constant.

    The commutator for position and momentum in quantum mechanics (QM) is standard knowledge for a student of QM, thus the result is given here without proof:
    [tex] \left[ x,p \right] = \left[ x, -i\hbar \frac{d}{dx} \right] = i\hbar . [/tex]
    To prove this, act with this commutator on a test function.

    Now let us consider the variance in QM:
    [tex] (\Delta x ) ^2 = \int \psi^*(\Delta x ) ^2\psi dx [/tex]
    [tex]( \Delta x ) = x - <x> [/tex]
    Now we have this very nice relation:
    [tex]\left[ \Delta x,\Delta p \right] = \left[ x,p\right] = i \hbar, [/tex] as an exercise, show this.

    Now substitute [itex] a(x) = A\psi (x) [/itex] and [itex] b(x) = B\psi (x)[/itex] in the Schwarz inequality:

    [tex]\int(A\psi)^*(A\psi)dx\int(B\psi)^*(B\psi)dx = <A^2><B^2> \geq \left| \int (A\psi)^*(B\psi) dx \right|^2 = \left| \int \psi^*(A(B\psi)) dx \right|^2 [/tex]

    [tex] \int \psi^* AB \psi dx = \frac{1}{2}\int \psi^* (AB+BA) \psi dx + \frac{1}{2}\int \psi^* (AB-BA) \psi dx [/tex]

    We have:
    [tex] <A^2><B^2> \geq \left| \frac{1}{2}\int \psi^* (AB+BA) \psi dx + \frac{1}{2}\int \psi^* (AB-BA) \psi dx \right|^2 \Rightarrow [/tex]

    Standard algebra: [itex]|a+b|^2 \geq |a|^2 + |b|^2[/itex]
    [tex] <A^2><B^2> \geq \frac{1}{4}\left| \int \psi^* (AB+BA) \psi dx \right|^2+ \frac{1}{4}\left|\int \psi^* (AB-BA) \psi dx \right|^2 [/tex]

    The first term on the right hand side is a number greater than zero (it is equal to the integral [itex] I[/itex] which is real, real number squared is a number greater than zero). So the lower limit is:
    [tex] <A^2><B^2> \geq \frac{1}{4}\left|\int \psi^* \left[ A,B \right] \psi dx \right|^2 [/tex]

    Make substituion [itex] A \rightarrow \Delta x [/itex] and [itex] B \rightarrow \Delta p [/itex] and use the fact that the expectation value of variance is equal to the variance, and that wavefunctions are normalized to unity:

    [tex](\Delta x)^2(\Delta p)^2 \geq \frac{1}{4} \left|\int \psi^* \left[ \Delta x,\Delta p \right] \psi dx \right|^2 [/tex]
    [tex] (\Delta x)^2(\Delta p)^2 \geq \frac{1}{4} (-i\hbar)(i\hbar)\cdot 1 = \hbar ^2/4 [/tex]

    [tex] \Delta x\Delta p \geq \hbar/2 [/tex]

    Consider Heisenberg Uncertainty Principle - Derivation (II) for same result obtained in bra-ket notation.

    We can also choose to perform this uncertainty with any two operators. It will become interesting if we consider two operators which does not commute. e.g let us consider the angular momentum operators, which have the following commutator:

    [tex] [ L_x, L_y] = i\hbar L_z [/tex]

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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