Comprehensive Study Notes

We now develop the theory of quantum mechanics in a more general way than previously. We begin by writing the one-particle, one-dimensional, time-independent Schrödinger equation (1.19) in the form

\(
\begin{equation}
\left[-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+V(x)\right] \psi(x)=E \psi(x) \tag{3.1}
\end{equation}
\)

The entity in brackets in (3.1) is an operator. Equation (3.1) suggests that we have an energy operator, which, operating on the wave function, gives us the wave function back again, but multiplied by an allowed value of the energy. We therefore discuss operators.

An operator is a rule that transforms a given function into another function. For example, let $\hat{D}$ be the operator that differentiates a function with respect to $x$. We use a circumflex to denote an operator. Provided $f(x)$ is differentiable, the result of operating on $f(x)$ with $\hat{D}$ is $\hat{D} f(x)=f^{\prime}(x)$. For example, $\hat{D}\left(x^{2}+3 e^{2 x}\right)=2 x+6 e^{2 x}$. If $\hat{3}$ is the operator that multiplies a function by 3 , then $\hat{3}\left(x^{2}+3 e^{x}\right)=3 x^{2}+9 e^{x}$. If tan is the operator that takes the tangent of a function, then application of $\tan$ to the function $x^{2}+1$ gives $\tan \left(x^{2}+1\right)$. If the operator $\hat{A}$ transforms the function $f(x)$ into the function $g(x)$, we write $\hat{A} f(x)=g(x)$.

We define the sum and the difference of two operators $\hat{A}$ and $\hat{B}$ by

\(
\begin{align}
& (\hat{A}+\hat{B}) f(x) \equiv \hat{A} f(x)+\hat{B} f(x) \tag{3.2}\
& (\hat{A}-\hat{B}) f(x) \equiv \hat{A} f(x)-\hat{B} f(x)
\end{align}
\)

For example, if $\hat{D} \equiv d / d x$, then

\(
(\hat{D}+\hat{3})\left(x^{3}-5\right) \equiv \hat{D}\left(x^{3}-5\right)+\hat{3}\left(x^{3}-5\right)=3 x^{2}+\left(3 x^{3}-15\right)=3 x^{3}+3 x^{2}-15
\)

An operator can involve more than one variable. For example, the operator $\partial^{2} / \partial x^{2}+\partial^{2} / \partial y^{2}$ has the following effect:

\(
\left(\partial^{2} / \partial x^{2}+\partial^{2} / \partial y^{2}\right) g(x, y)=\partial^{2} g / \partial x^{2}+\partial^{2} g / \partial y^{2}
\)

The product of two operators $\hat{A}$ and $\hat{B}$ is defined by

\(
\begin{equation}
\hat{A} \hat{B} f(x) \equiv \hat{A}[\hat{B} f(x)] \tag{3.3}
\end{equation}
\)

In other words, we first operate on $f(x)$ with the operator on the right of the operator product, and then we take the resulting function and operate on it with the operator on the left of the operator product. For example, $\hat{3} \hat{D} f(x)=\hat{3}[\hat{D} f(x)]=\hat{3} f^{\prime}(x)=3 f^{\prime}(x)$.

The operators $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ may not have the same effect. Consider, for example, the operators $d / d x$ and $\hat{x}$ (where $\hat{x}$ means multiplication by $x$ ):

\(
\begin{gather}
\hat{D} \hat{x} f(x)=\frac{d}{d x}[x f(x)]=f(x)+x f^{\prime}(x)=(\hat{1}+\hat{x} \hat{D}) f(x) \tag{3.4}\
\hat{x} \hat{D} f(x)=\hat{x}\left[\frac{d}{d x} f(x)\right]=x f^{\prime}(x)
\end{gather}
\)

Thus $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ are different operators in this case.
We can develop an operator algebra as follows. Two operators $\hat{A}$ and $\hat{B}$ are said to be equal if $\hat{A} f=\hat{B} f$ for all functions $f$. Equal operators produce the same result when they operate on a given function. For example, (3.4) shows that

\(
\begin{equation}
\hat{D} \hat{x}=1+\hat{x} \hat{D} \tag{3.5}
\end{equation}
\)

The operator $\hat{1}$ (multiplication by 1) is the unit operator. The operator $\hat{0}$ (multiplication by 0 ) is the null operator. We usually omit the circumflex over operators that are simply multiplication by a constant. We can transfer operators from one side of an operator equation to the other (Prob. 3.7). Thus (3.5) is equivalent to $\hat{D} \hat{x}-\hat{x} \hat{D}-1=0$, where circumflexes over the null and unit operators were omitted.

Operators obey the associative law of multiplication:

\(
\begin{equation}
\hat{A}(\hat{B} \hat{C})=(\hat{A} \hat{B}) \hat{C} \tag{3.6}
\end{equation}
\)

The proof of (3.6) is outlined in Prob. 3.10. As an example, let $\hat{A}=d / d x, \hat{B}=\hat{x}$, and $\hat{C}=3$. Using (3.5), we have

\(
\begin{array}{ll}
(\hat{A} \hat{B})=\hat{D} \hat{x}=1+\hat{x} \hat{D}, & {[(\hat{A} \hat{B}) \hat{C}] f=(1+\hat{x} \hat{D}) 3 f=3 f+3 x f^{\prime}} \
(\hat{B} \hat{C})=3 \hat{x}, & {[\hat{A}(\hat{B} \hat{C})] f=\hat{D}(3 x f)=3 f+3 x f^{\prime}}
\end{array}
\)

A major difference between operator algebra and ordinary algebra is that numbers obey the commutative law of multiplication, but operators do not necessarily do so; $a b=b a$ if $a$ and $b$ are numbers, but $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ are not necessarily equal operators. We define the commutator $[\hat{A}, \hat{B}]$ of the operators $\hat{A}$ and $\hat{B}$ as the operator $\hat{A} \hat{B}-\hat{B} \hat{A}$ :

\(
\begin{equation}
[\hat{A}, \hat{B}] \equiv \hat{A} \hat{B}-\hat{B} \hat{A} \tag{3.7}
\end{equation}
\)

If $\hat{A} \hat{B}=\hat{B} \hat{A}$, then $[\hat{A}, \hat{B}]=0$, and we say that $\hat{A}$ and $\hat{B}$ commute. If $\hat{A} \hat{B} \neq \hat{B} \hat{A}$, then $\hat{A}$ and $\hat{B}$ do not commute. Note that $[\hat{A}, \hat{B}] f=\hat{A} \hat{B} f-\hat{B} \hat{A} f$. Since the order in which we apply the operators 3 and $d / d x$ makes no difference, we have

\(
\left[\hat{3}, \frac{d}{d x}\right]=\hat{3} \frac{d}{d x}-\frac{d}{d x} \hat{3}=0
\)

From Eq. (3.5) we have

\(
\begin{equation}
\left[\frac{d}{d x}, \hat{x}\right]=\hat{D} \hat{x}-\hat{x} \hat{D}=1 \tag{3.8}
\end{equation}
\)

The operators $d / d x$ and $\hat{x}$ do not commute.

EXAMPLE

Find $\left[z^{3}, d / d z\right]$.
To find $\left[z^{3}, d / d z\right]$, we apply this operator to an arbitrary function $g(z)$. Using the commutator definition (3.7) and the definitions of the difference and product of two operators, we have

\(
\begin{aligned}
{\left[z^{3}, d / d z\right] g=\left[z^{3}(d / d z)-(d / d z) z^{3}\right] g } & =z^{3}(d / d z) g-(d / d z)\left(z^{3} g\right) \
& =z^{3} g^{\prime}-3 z^{2} g-z^{3} g^{\prime}=-3 z^{2} g
\end{aligned}
\)

Deleting the arbitrary function $g$, we get the operator equation $\left[z^{3}, d / d z\right]=-3 z^{2}$.

EXERCISE Find $\left[d / d x, 5 x^{2}+3 x+4\right]$. (Answer: $10 x+3$.)

The square of an operator is defined as the product of the operator with itself: $\hat{B}^{2}=\hat{B} \hat{B}$. Let us find the square of the differentiation operator:

\(
\begin{aligned}
\hat{D}^{2} f(x) & =\hat{D}(\hat{D} f)=\hat{D} f^{\prime}=f^{\prime \prime} \
\hat{D}^{2} & =d^{2} / d x^{2}
\end{aligned}
\)

As another example, the square of the operator that takes the complex conjugate of a function is equal to the unit operator, since taking the complex conjugate twice gives the original function. The operator $\hat{B}^{n}(n=1,2,3, \ldots)$ is defined to mean applying the operator $\hat{B} n$ times in succession.

It turns out that the operators occurring in quantum mechanics are linear. $\hat{A}$ is a linear operator if and only if it has the following two properties:

\(
\begin{equation}
\hat{A}[f(x)+g(x)]=\hat{A} f(x)+\hat{A} g(x) \tag{3.9}
\end{equation}
\)

\(
\begin{equation}
\hat{A}[c f(x)]=c \hat{A} f(x) \tag{3.10}
\end{equation}
\)

where $f$ and $g$ are arbitrary functions and $c$ is an arbitrary constant (not necessarily real). Examples of linear operators include $\hat{x}^{2}, d / d x$, and $d^{2} / d x^{2}$. Some nonlinear operators are cos and ()$^{2}$, where ()$^{2}$ squares the function it acts on.

EXAMPLE

Is $d / d x$ a linear operator? Is $\sqrt{ }$ a linear operator?
We have

\(
\begin{gathered}
(d / d x)[f(x)+g(x)]=d f / d x+d g / d x=(d / d x) f(x)+(d / d x) g(x) \
(d / d x)[c f(x)]=c d f(x) / d x
\end{gathered}
\)

so $d / d x$ obeys (3.9) and (3.10) and is a linear operator. However,

\(
\sqrt{f(x)+g(x)} \neq \sqrt{f(x)}+\sqrt{g(x)}
\)

so $\sqrt{ }$ does not obey (3.9) and is nonlinear.
EXERCISE Is the operator $x^{2} \times$ (multiplication by $x^{2}$ ) linear? (Answer: Yes.)

Useful identities in linear-operator manipulations are

\(
\begin{equation}
(\hat{A}+\hat{B}) \hat{C}=\hat{A} \hat{C}+\hat{B} \hat{C} \tag{3.11}
\end{equation}
\)

\(
\begin{equation}
\hat{A}(\hat{B}+\hat{C})=\hat{A} \hat{B}+\hat{A} \hat{C} \tag{3.12}
\end{equation}
\)

EXAMPLE

Prove the distributive law (3.11) for linear operators.
A good way to begin a proof is to first write down what is given and what is to be proved. We are given that $\hat{A}, \hat{B}$, and $\hat{C}$ are linear operators. We must prove that $(\hat{A}+\hat{B}) \hat{C}=\hat{A} \hat{C}+\hat{B} \hat{C}$.

To prove that the operator $(\hat{A}+\hat{B}) \hat{C}$ is equal to the operator $\hat{A} \hat{C}+\hat{B} \hat{C}$, we must prove that these two operators give the same result when applied to an arbitrary function $f$. Thus we must prove that

\(
[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A} \hat{C}+\hat{B} \hat{C}) f
\)

We start with $[(\hat{A}+\hat{B}) \hat{C}] f$. This expression involves the product of the two operators $\hat{A}+\hat{B}$ and $\hat{C}$. The operator-product definition (3.3) with $\hat{A}$ replaced by $\hat{A}+\hat{B}$ and $\hat{B}$ replaced by $\hat{C}$ gives $[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A}+\hat{B})(\hat{C} f)$. The entity $\hat{C} f$ is a function, and use of the definition (3.2) of the sum $\hat{A}+\hat{B}$ of the two operators $\hat{A}$ and $\hat{B}$ gives $(\hat{A}+\hat{B})(\hat{C} f)=\hat{A}(\hat{C} f)+\hat{B}(\hat{C} f)$. Thus

\(
[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A}+\hat{B})(\hat{C} f)=\hat{A}(\hat{C} f)+\hat{B}(\hat{C} f)
\)

Use of the operator-product definition (3.3) gives $\hat{A}(\hat{C} f)=\hat{A} \hat{C} f$ and $\hat{B}(\hat{C} f)=\hat{B} \hat{C} f$. Hence

\(
\begin{equation}
[(\hat{A}+\hat{B}) \hat{C}] f=\hat{A} \hat{C} f+\hat{B} \hat{C} f \tag{3.13}
\end{equation}
\)

Use of the operator-sum definition (3.2) with $\hat{A}$ replaced by $\hat{A} \hat{C}$ and $\hat{B}$ replaced by $\hat{B} \hat{C}$ gives $(\hat{A} \hat{C}+\hat{B} \hat{C}) f=\hat{A} \hat{C} f+\hat{B} \hat{C} f$, so (3.13) becomes

\(
[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A} \hat{C}+\hat{B} \hat{C}) f
\)

which is what we wanted to prove. Hence $(\hat{A}+\hat{B}) \hat{C}=\hat{A} \hat{C}+\hat{B} \hat{C}$.
Note that we did not need to use the linearity of $\hat{A}, \hat{B}$, and $\hat{C}$. Hence (3.11) holds for all operators. However, (3.12) holds only if $\hat{A}$ is linear (see Prob. 3.17).

EXAMPLE

Find the square of the operator $d / d x+\hat{x}$.
To find the effect of $(d / d x+\hat{x})^{2}$, we apply this operator to an arbitrary function $f(x)$. Letting $\hat{D} \equiv d / d x$, we have

\(
\begin{aligned}
(\hat{D}+\hat{x})^{2} f(x)= & (\hat{D}+\hat{x})[(\hat{D}+x) f]=(\hat{D}+\hat{x})\left(f^{\prime}+x f\right) \
= & f^{\prime \prime}+f+x f^{\prime}+x f^{\prime}+x^{2} f=\left(\hat{D}^{2}+2 \hat{x} \hat{D}+\hat{x}^{2}+1\right) f(x) \
& (\hat{D}+\hat{x})^{2}=\hat{D}^{2}+2 \hat{x} \hat{D}+\hat{x}^{2}+1
\end{aligned}
\)

Let us repeat this calculation, using only operator equations:

\(
\begin{aligned}
(\hat{D}+\hat{x})^{2} & =(\hat{D}+\hat{x})(\hat{D}+\hat{x})=\hat{D}(\hat{D}+\hat{x})+\hat{x}(\hat{D}+\hat{x}) \
& =\hat{D}^{2}+\hat{D} \hat{x}+\hat{x} \hat{D}+\hat{x}^{2}=\hat{D}^{2}+\hat{x} \hat{D}+1+\hat{x} \hat{D}+\hat{x}^{2} \
& =\hat{D}^{2}+2 x \hat{D}+x^{2}+1
\end{aligned}
\)

where (3.11), (3.12), and (3.5) have been used and the circumflex over the operator "multiplication by $x$ " has been omitted. Until you have become thoroughly experienced with operators, it is safest when doing operator manipulations always to let the operator operate on an arbitrary function $f$ and then delete $f$ at the end.

EXERCISE Find $\left(d^{2} / d x^{2}+x\right)^{2}$. (Answer: $d^{4} / d x^{4}+2 x d^{2} / d x^{2}+2 d / d x+x^{2}$.)