11.3.2: Uncertainty in Quantum Mechanics
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In order to bring this into quantum mechanics, we already know how to calculate the average ⟨a⟩, which we call the “expectation value”. If the state of the system is |ψ⟩ and the operator corresponding to the observable a is ˆA, then
⟨a⟩=⟨ψ|ˆA|ψ⟩
Similarly, now that we recognize that we can interpret ˆA2 as just applying the operator ˆA twice, we can calculate ⟨a2⟩:
⟨a2⟩=⟨ψ|ˆA2|ψ⟩
For example, let’s consider the state |ψ⟩=|+z⟩ and the observable spin-z. We expect the uncertainty here to be zero, because we know exactly what we’ll get if we measure spin-z. Let’s see if it works out that way:
⟨sz⟩=⟨ψ|ˆSz|ψ⟩=ℏ2[10][100−1][10]=ℏ2[10][10]=ℏ2
As expected, the expectation value for spin-z is +ℏ/2. For the other part:
⟨s2z⟩=⟨+z|ˆSzˆSz|−z⟩=ℏ24[10][100−1][100−1][10]=ℏ24[10][100−1][10]=ℏ24[10][10]=ℏ24
If we take the difference ⟨sz2⟩−⟨sz⟩2, we get ℏ2/4−ℏ2/4=0, as expected.
What if we want to know the uncertainty on Sx for this state?
⟨sx⟩=⟨+z|ˆSx|+z⟩=ℏ2[10][0110][10]=ℏ2[10][01]=0
If the system is in the state |+z⟩, we know that we have a 50% chance each for finding spin-x to be +ℏ/2 or −ℏ/2. Thus, it’s no surprise that the average value of spin-x is zero, even though zero isn’t a value we might measure. To figure out the variance:
⟨s2x⟩=⟨+z|ˆSxˆSx|+z⟩=ℏ24[10][0110][0110][10]=ℏ24[10][0110][01]=ℏ24[10][10]=ℏ24
Thus, in this case, the formal uncertainty Δsx on the x-spin is ℏ/2.