1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the uncertainty principle

  1. Jul 24, 2014 #1
    Definition/Summary

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

    Equations

    [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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted