BrendanM
Layra-chan
(Δa)(Δb) >= |<Ψ|(AB - BA)|Ψ>|/2
I'm going to pretend I understand that and nod knowingly.
P.S. WTF does that mean?
For a given property a, Δa is the amount of uncertainty in measurement/knowledge; it's the average amount of how much the measured value differs from the expected value. Thus (Δa)(Δb) means the amount of uncertainty there is in the measurement of property a times the amount of uncertainty there is in the measurement of property b. In some sense, this uncertainty in measurement can mean an uncertainty in actual existence, i.e. if Δa is positive, then a doesn't have an actual, definite value but rather is spread out amongst many possible values.
The right side is the tricky part. Warning: long read ahead that requires calculus. See next post for shorter, less technical summary:
In quantum mechanics, the state of the system is described by a wavefunction, usually written as |Ψ>. We're going to consider it to be a function of position that returns a complex value. Due to notational difficulties, as a function of position x the wavefunction is usually written as Ψ(x), but the | > notation comes in handy for formal manipulations.
We assume that the integral of |Ψ|²(x) as x goes from negative infinity to infinity, is 1. Nothing interesting or useful happens if we don't assume this, and it's convenient.
Anyway, a given property
a of the system, such as the position of a given particle, is gotten by applying a corresponding operator
A to the wavefunction Ψ(x) to get a new wavefunction (
AΨ)(x), also written
A|Ψ>.
Don't worry about what "operator" means exactly; it's just a function that takes a wavefunction to a wavefunction or to a number of wavefunctions. Just like functions on, say, the reals, we can add two operators by adding their outputs, and we can compose them by applying the second operator to the output of the first operator, assuming the first operator spits out only one wavefunction.
If the property is well-defined for the system, then
A|Ψ> is a multiple of |Ψ>; specifically, we get that
A|Ψ> =
a|Ψ> where
a is the measured value. In other words, there is some fixed value of
a such that for all x, (
AΨ)(x) =
aΨ(x)
Even if the property is not well-defined for the system, we can still take the average by calculating <Ψ|
A|Ψ>, which just means the integral of Ψ*(x)(
AΨ)(x) as x goes from negative infinity to positive infinity. Note that if (
AΨ)(x) =
aΨ(x), then <Ψ|
A|Ψ> =
a since
a is a constant and the integral of Ψ*(x)Ψ(x) is 1.
Now suppose we want to know the values of the product of two different properties
a and
b with corresponding operators
A and
B. What can we do? Well, we can find <Ψ|
AB|Ψ>. But we can also find <Ψ|
BA|Ψ>. At first glance they look the same, right?
Nope. Operators don't necessarily commute, i.e. A(BΨ) doesn't necessarily equal B(AΨ). The simplest example is position and momentum. In one dimension, the position operator X takes a wavefunction |Ψ> and multiplies it pointwise by the position x, i.e.
(XΨ)(x) = xΨ(x)
while the momentum operator P takes the derivative with respect to x and then multiplies by -ih, i.e.
(PΨ)(x) = -ih(dΨ/dx)(x)
Never mind the strange form of P, that follows from deeper principles. h is a positive constant (the reduced Planck constant) whose exact value is unimportant for this discussion. The important part is that first applying the position operator followed by the momentum operator is different from first applying the momentum operator followed by the position operator, i.e.
P(XΨ) = -ih(d/dx)(xΨ(x)) = -ih(Ψ(x)+x(dΨ/dx)(x)) by the product rule for differentiation
X(PΨ) = -ih(x(dΨ/dx)(x))
(P(XΨ))(x) - (X(PΨ))(x) = ((PX-XP)Ψ)(x) = -ihΨ(x)
Thus we get that <Ψ|PX-XP|Ψ> = -ih, so |Ψ|PX-XP|Ψ>| = h
So the formula says that (Δx)(Δp) >= h/2, i.e. the amount of uncertainty in the position times the amount of uncertainty in the momentum is at least h/2, where h is positive. This says that if you know the position of a particle exactly, then Δp is infinity and you know nothing about the momentum; conversely, if you know the momentum of the particle exactly, then Δx is infinity and the particle is smeared across the entire universe. A happier medium is that you can't know either position or momentum exactly, with a bound on your certainty given by the formula above.
