Michael Fowler 12/ 9/06
As we discussed in the Linear Algebra lecture, if two physical variables correspond to commuting Hermitian operators, they can be diagonalized simultaneously—that is, they have a common set of eigenstates. In these eigenstates both variables have precise values at the same time, there is no “Uncertainty Principle” requiring that as we know one of them more accurately, we increasingly lose track of the other. For example, the energy and momentum of a free particle can both be specified exactly. More interesting examples will appear in the sections on angular momentum and spin.
But if two operators do not commute, in general one cannot specify both values precisely. Of course, such operators could still have some common eigenvectors, but the interesting case arises in attempting to measure A and B simultaneously for a state in which the commutator has a nonzero expectation value, .
Our task here is to give a quantitative analysis of how accurately noncommuting variables can be measured together. We found earlier using a semi-quantitative argument that for a free particle, at best. To improve on that result, we need to be precise about the uncertainty in a state
We define as the root mean square deviation:
To make the equations more compact, we define by
(We’ll put a caret (a hat) on the to remind ourselves it’s an operator—and, of course, it’s a Hermitian operator, like A.) We also drop the bra and ket, on the understanding that this whole argument is for a particular state. Now
Introduce an operator B in exactly similar fashion, having the property that .
The quantitative measure of how the combined “uncertainty” of measuring two variables relates to their lack of commutativity is most simply presented as a
(Remark: remember that for A, B Hermitian, is antiHermitian: so is real! To make this clear, take adjoints: if H is hermitian then so is real. If K is anti Hermitian, then from which is pure imaginary.)
Proof of the Theorem:
Using Schwartz’s inequality
The operator is neither Hermitian nor antiHermitian. To evaluate the mod squared of its expectation value, we break the amplitude into real and imaginary parts:
(The first term on the right-hand side is the expectation value of a Hermitian matrix, and so is real, the second term is the expectation value of an antiHermitian matrix, so is pure imaginary.)
It follows immediately that
For a particle in one dimension denote
(It important in that last step to understand that the operator operates on everything to its right, and, as we are always finding matrix elements of operators, there will be a following ket it operates on, so )
We conclude that
Question: Is there a wavefunction for which this inequality becomes an equality?
That would require which can only be true if the two vectors are parallel, or, written explicitly,
Actually , that’s not enough: we also need to be zero. (Look at the equation above giving in terms of its real and imaginary parts, and how we used it to establish the inequality.)
Writing as we find
so this will be zero if and only if is pure imaginary.
Turning to the differential equation, we first take the simplest case where and are both zero. The first requirement just sets the origin, but the second stipulates that our wave function has no net momentum.
For this simple case, becomes
and recalling that is pure imaginary, this is a Gaussian wave packet! It is straightforward to check that the solution with and nonzero is
where is real, and C is the usual Gaussian normalization constant.
Exercise: check this.
The conclusion is then that the Gaussian wave packet gives the optimum case for minimizing the joint uncertainties in position and momentum.
Note that the condition does not mean that is an eigenstate of either , but it is an eigenstate of the nonHermitian operator , with eigenvalue zero. We shall soon see that this nonHermitian operator and its adjoint play important roles in the quantum mechanics of the simple harmonic oscillator.