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What is the uncertainty principle

  1. Jul 24, 2014 #1

    This is a derivation of the Uncertainty Principle based on the properties of non-commuting Hermitian operators.


    [tex]\langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \geq \frac{1}{4} |\langle [A,B] \rangle | ^2 [/tex]

    [tex]\langle (\Delta x_i)^2 \rangle \langle (\Delta p_i)^2 \rangle \geq \frac{1}{4} \hbar ^2 [/tex]

    Extended explanation

    Let A, B be a pair of operators. We define [itex]\Delta A \equiv A - \langle A \rangle I [/itex], where the expectation value of A with respect to some state [itex]|a \rangle [/itex], is defined as [itex]\langle A \rangle = \langle a | A | a \rangle [/itex]. This number tells you what A will be measured as, on average, over several repeated measurements performed on the system, when prepared identically.

    Now, we define an important quantity: the variance or mean square deviation, which is [itex]\langle ( \Delta A ) ^2 \rangle [/itex]. This quantity is no different from the variance in any statistical collection of data. Plugging in from above

    [tex]\langle ( \Delta A ) ^2 \rangle = \langle (A - \langle A \rangle I)^2 \rangle = \langle A^2 \rangle - \langle A \rangle ^2 ~~~-~(1)[/tex]

    Let [itex]|x \rangle [/itex] be any arbitrary (but normalized) state ket. Let

    [tex]|a \rangle = \Delta A ~ |x \rangle [/tex]
    [tex]|b \rangle = \Delta B ~ |x \rangle [/tex]

    First we apply the Cauchy-Schwarz inequality (which is essentially a result that is two steps removed from saying that the length of a vector is a positive, real number): [itex] \langle a |a \rangle \langle b |b \rangle \geq | \langle a |b \rangle |^2 [/itex], to the above kets (keeping in mind that [itex]\Delta A~, ~\Delta B [/itex] are Hermitian), giving

    [tex]\langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \geq |\langle \Delta A \Delta B \rangle | ^2 ~~~-~(2)[/tex]

    Next we write

    [tex]\Delta A \Delta B = \frac{1}{2}(\Delta A \Delta B - \Delta B \Delta A) + \frac{1}{2}(\Delta A \Delta B + \Delta B \Delta A) = \frac{1}{2}[\Delta A, \Delta B] + \frac{1}{2}\{ \Delta A, \Delta B \} ~~~-~(3) [/tex]

    Now, the commutator

    [tex][\Delta A,~ \Delta B] = [A - \langle A \rangle I,~B - \langle B \rangle I] = [A,B] ~~~-~(4)[/tex]

    And notice that [itex] [A,B] [/itex] is anti-Hermitian, giving it a purely imaginary expectation value. On the other hand, the anti-commutator [itex]\{ \Delta A,~ \Delta B \} [/itex] is clearly Hermitian, and so, has a real expectation. Thus

    [tex]\langle \Delta A \Delta B \rangle = \frac{1}{2}\langle [A,B] \rangle + \frac{1}{2} \langle \{ \Delta A,~ \Delta B \} \rangle ~~~-~(5)[/tex]

    Since the terms on the RHS are merely the real and imaginary parts of the expectation on the LHS, we have

    [tex]| \langle \Delta A \Delta B \rangle |^2 = \frac{1}{4}| \langle [ A,B] \rangle |^2 + \frac{1}{4} | \langle \{ \Delta A,~ \Delta B\} \rangle |^2 \geq \frac{1}{4}| \langle [A,B] \rangle |^2~~~-~(6)[/tex]

    Using the result of (6) in (2) gives :

    [tex]\langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \geq \frac{1}{4} |\langle [A,B] \rangle | ^2 ~~~-~(7)[/tex]

    The above equation (7), is the most general form of the Uncertainty Relation for a pair of hermitian operators. So far, it is nothing more than a statement of a particular property of certain specifically constructed hermitian matrices.

    Notice that if the operators A, B commute (ie: [A,B] = 0), then the product of the variances vanish, and there is no uncertanity in measuring their observables simultaneously. It is only in the case of non-commuting (or incompatible) operators, that you see the more popular form of the Uncertainty Principle, where the product of the variances does not vanish.

    Specifically, in the case where [itex]A = \hat{x_i}~,~~B = \hat{p_i} [/itex], we use the commutation relation:

    [tex] [\hat{x_i},\hat{p_i}] = i \hbar ~~~-~(8)[/tex]

    This equation follows from the definition of the quantum mechanical momentum operator, which is constructed upon the following two observations:

    (i) In classical mechanics, momentum is the generator of infintesimal translations. The infinitesimal translation operator, [itex]\tau (d \mathbf{x}) [/itex], defined by [itex]\tau (d \mathbf{x}) |\mathbf{x} \rangle \equiv |\mathbf{x} + d \mathbf{x} \rangle [/itex] can be written as

    [tex] \tau (d \mathbf{x}) = I - i\mathbf{K} \cdot d \mathbf{x} [/tex]

    (ii) K is an operator with dimension length -1, and hence, can be written as [itex] \mathbf{K} = \mathbf{p} / [action] [/itex]. The choice of this universal constant with dimensions of action (energy*time) comes from the de broglie observation [itex]k = p/ \hbar [/itex]. So, writing [itex] \tau (d \mathbf{x}) = I - i\mathbf{p} \cdot d \mathbf{x} /\hbar [/itex] leads to the expected commutation relation , [itex][\hat{x_i}, \hat{p_i} ] = i \hbar [/itex].

    Plugging this into (7) gives the correct expression for the HUP:

    [tex]\langle (\Delta x_i)^2 \rangle \langle (\Delta p_i)^2 \rangle \geq \frac{1}{4} \hbar ^2 [/tex]

    It is this expression that is often popularized in the (somewhat misleading) short-hand: [itex]\Delta x \Delta p \geq \hbar/2[/itex]

    * 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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