# What is expected value

1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

A short introduction to expectation value is given, both for discrete and continuous cases.

Equations

For discrete probability distributions,

$$<Q> \ = \ \sum _n Q_n p_n$$

For continuous distributions specified by a normalized, real space wave-function $\psi(x)$,

$$< Q > = \int _{\text{All space}} \psi^*(x)Q(x)\psi(x) dx$$

Extended explanation

NOTATION:

The notation < > comes from statistics, so it is a general notation which QM scientists borrowed.

DEFINITIONS:

The expectation value of an observable associated with an operator $Q$ is defined as:

$$<Q> = \sum _n Q_n p_n$$

in the case of a discrete spectrum, where $Q_n$ is the eigenvalue of Q for a state labeled by the index n, and $p_n$ is the probability of measuring the system in this state.

DISCRETE DISTRIBUTIONS:

Variance in statistics, discrete case:
$$(\Delta A ) ^2 = \sum _n (A_n - <A>)^2 p_n ,$$
$$\sum _n p_n = 1 ,$$
$$<A> = \sum _n A_nP_n ,$$
$$<<A>> = <A>$$
$<A>$ is just a number, we can thus show that:
$$(\Delta A ) ^2 = <A^2> + <A>^2$$
and
$$<(\Delta A ) ^2> = (\Delta A ) ^2.$$ as an exercise, show this.

where $\sum _n p_n = 1$ and $A_n$ is the outcome of the n'th value.

EXAMPLE:

As an exercise, let's find the expectation value <D>, of the outcome of rolling dice:

$$<D> = 1 \cdot \dfrac{1}{6} + 2 \cdot \dfrac{1}{6} + 3 \cdot \dfrac{1}{6} + 4 \cdot \dfrac{1}{6} + 5 \cdot \dfrac{1}{6} + 5 \cdot \dfrac{1}{6} = \dfrac{7}{2}$$
since each value has the equal probability of $1/6 .$

CONTINUOUS DISTRIBUTIONS:

Now this was for the discrete case, in the continous case:
$$< Q > = \int _{\text{All space}} f(x)Q(x)f(x) dx$$
where $f^2(x)$ is the probability density distribution : $\int f^2(x) dx = 1$.

That was if the probability density distribution is real, for complex valued (such as quantum mechanical wave functions):
$$< Q > = \int _{\text{All space}} \psi^*(x)Q(x)\psi(x) dx$$
$\int |\psi (x)|^2 dx = 1$.

EXAMPLES:

Position:
$$< x > = \int _{\text{All space}} \psi^*(x)x\psi(x) dx = \int x|\psi (x)|^2 dx$$

Momentum:
$$< p > = \int _{\text{All space}} \psi^*(x)(-i\hbar\dfrac{d}{dx})\psi(x) dx$$

Now the variance is:
$$\Delta Q ^2 = <(Q - <Q>)^2> = <Q^2> - <Q>^2$$

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