📂Quantum Mechanics

Expectation Value in Quantum Mechanics

Definition

The expectation value of a normalized wavefunction $\psi$ for an operator $A$ is defined as follows.

$$\braket{A} =\int_{-\infty}^{\infty} \psi^{\ast}A\psi dx$$

Explanation

To put it simply, it is the same expectation value that you learned in high school statistics. When studying quantum mechanics, if you find it difficult to understand the expectation value, the difficulties can be broadly classified into two types. The first is the difficulty in understanding the definition itself and the second is the question of why the formula is expressed in that way.

What is the expectation value?

As you know, when you throw a dice, the expectation value of the dice roll is $(1+2+3+4+5+6)*\dfrac{ 1 }{ 6 }=3.5$. It means that if you throw the dice many times, the average will be close to $3.5$. Similarly, the expectation value of the momentum of a particle is the average of the measured momentum values.

It’s important to note that this is not about measuring one particle multiple times because, after measurement, the superposed wavefunction collapses to a single value, always yielding the same result. Therefore, the expectation value of a particle’s momentum is the expectation value when measuring particles of the same type under the same conditions.

To understand more easily, let’s compare particles to people and momentum to weight. The expectation value of a particle is obtained not by repeatedly measuring the same person but by measuring the weights of multiple people who fit the criteria of 20-year-old males in Korea. Just as measuring the same person’s weight multiple times yields the same result, repeated measurements of the same particle yield the same result.

Why does the formula look like that?

Let’s review the definition and notation.

$$\braket{A} =\int_{-\infty}^{\infty} \psi^{\ast}A\psi dx \tag{1}$$

Although it may look different from the expectation value you learned in high school mathematics, it is essentially the same. If the probability density function of a continuous random variable $X$ is $f(x)$, then its expectation value is

$$E(X)=\int_{-\infty}^{\infty}xf(x)dx \tag{2}$$

Now let’s see why these two are the same. Since the probability density of $\psi$ is $| \psi|^{2}=\psi^{\ast}\psi$, writing $\braket{ A}$ as $(2)$ gives

$$\begin{align*} \braket{A}& = \int_{-\infty}^{\infty}A|\psi|^{2}dx \\ &= \int_{-\infty}^{\infty}A\psi^{\ast}\psi dx \\ &= \int_{-\infty}^{\infty}\psi^{\ast}A\psi dx \end{align*}$$

This should resolve the question of why the expectation value formula looks the way it does. However, the question remains as to why the notation is $\psi^{\ast}A\psi$ instead of $A\psi^{\ast}\psi$ or $\psi A \psi^{\ast}$. In one dimension, the order of multiplication does not matter. However, in three dimensions, thinking of wave functions and operators as vectors and matrices respectively means the order of multiplication does matter. You probably know that matrices yield different results depending on the order of multiplication. For example, let’s assume we have an operator $A$ and a wave function $\psi$.

$$A=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}, \psi=\begin{pmatrix} \psi_{1} \\ \psi_{2} \\ \psi_{3} \end{pmatrix}$$

Then, the correct matrix multiplication is as follows.

$$\psi^{\ast} A \psi = \sum_{i,j = 1}^{3} \psi^{\ast}_{i} A_{ij} \psi_{j}$$

Variance

For a random variable $X$, if the expectation value is $\mu = E(X)$, then the variance of $X$ is defined as follows.

$$\sigma^{2} = E\left( (X - \mu)^{2} \right)$$

Similarly, the variance of an operator $A$ with respect to the wavefunction $\psi$ is defined as follows.

$$\sigma_{A}^{2} = \braket{(A - \braket{A})^{2}} = \braket{\psi | (A - \braket{A})^{2} | \psi} = \int_{-\infty}^{\infty} \psi^{\ast} (A - \braket{A})^{2} \psi dx$$