Commutations of Observables and Time Dependence of Expectation Values

_{Disclaimer: I lay no claim to the originality of the basic ideas expounded here. Its distinctness lies more in its scope and in the detail of its exposition.}

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Expectation Values of Observables

In quantum mechanical sense, what does the expectation value of an observable A mean?

The expectation value of an observable A in a particle state described by a wave function is defined as the average result of N, as it approaches a large number, of measurements of A performed on a particle in the same state .

The prescription for calculating the expectation value of a normalized wave function is known as

To show that this prescription is the same to the definition of the expectation value as an average, consider the following wavefunction,

which we have covered on the case on which is not an eigenfunction of the operator. We substitute this equation to the prescription of the expectation value,

This relation asserts that the expectation value of A in the state is the sum of all possible outcome , each weighted by the probability that in this state a measurement of A will yield the result .

Commuting observables and simultaneous observability

If the operators corresponding to two observables A and B have a common, complete set of eigenfunctions , where we have the i to label the corresponding eigenvalues of and j labels the corresponding eigenvalues of , we say that A and B can be simultaneously well defined. Another way of saying this is that they have a simultaneously observable.

How do we know that two operators are simultaneously observable?

The commutator of any pair of simultaneously well-defined observables A and B turns out to be zero. This is a sufficient condition.

In some sense, the commutator action on an arbitrary normalizable function gives zero. Lets show this.

We will use that fact that any such function can be written as a linear combination of the common eigenfunctions of , so that

The converse is also true, such that if and commute, that is, they are simultaneous eigenfunction.

Non-commuting observables and the uncertainty principle

If the commutator then the observables A and B cannot be simultaneously well defined.

The uncertainty in the observable A in the state is define by

We can show that the product of the uncertainties in two observables A and B is given by the inequality (general uncertainty relation)

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Proof:

Take note of this relation that we derive from previous section,

With this in mind, we set . So we have,

Using a similar line of reasoning we can show that,

Then the inequality becomes:

Now what is left is to show what

And so we have the following relation,

Example: Derive the uncertainty relation of the following operator: , then we have,

From this we can see that,

and so the uncertainty relation yields,

This implies that there is no 1D state that exists for which the product of the uncertainties of the position and momentum observable is smaller than . It is found that in a one dimensional harmonic oscillator in its ground state, , the uncertainty becomes an equality,

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Time dependence of Expectation Values

Use the following expression for the expectation value of an observable A ,

And also the time-dependent Schrodinger equation in the form,

From these equations, we can obtain an expression for the rate of change of the expectation value with time:

From this we can derive the following expression,

which is known as the Ehrenfest's theorem.