Probability plays a fundamental role in quantum mechanics. This section reviews the mathematics of probability.
There has been much controversy about the proper definition of probability. One definition is the following: If an experiment has $n$ equally probable outcomes, $m$ of which are favorable to the occurrence of a certain event $A$, then the probability that $A$ occurs is $m / n$. Note that this definition is circular, since it specifies equally probable outcomes when probability is what we are trying to define. It is simply assumed that we can recognize equally probable outcomes. An alternative definition is based on actually performing the experiment many times. Suppose that we perform the experiment $N$ times and that in $M$ of these trials the event $A$ occurs. The probability of $A$ occurring is then defined as
\(
\lim _{N \rightarrow \infty} \frac{M}{N}
\)
Thus, if we toss a coin repeatedly, the fraction of heads will approach $1 / 2$ as we increase the number of tosses.
For example, suppose we ask for the probability of drawing a heart when a card is picked at random from a standard 52 -card deck containing 13 hearts. There are 52 cards and hence 52 equally probable outcomes. There are 13 hearts and hence 13 favorable outcomes. Therefore, $m / n=13 / 52=1 / 4$. The probability for drawing a heart is $1 / 4$.
Sometimes we ask for the probability of two related events both occurring. For example, we may ask for the probability of drawing two hearts from a 52 -card deck, assuming we do not replace the first card after it is drawn. There are 52 possible outcomes of the first draw, and for each of these possibilities there are 51 possible second draws. We have $52 \cdot 51$ possible outcomes. Since there are 13 hearts, there are $13 \cdot 12$ different ways to draw two hearts. The desired probability is $(13 \cdot 12) /(52 \cdot 51)=1 / 17$. This calculation illustrates the theorem: The probability that two events $A$ and $B$ both occur is the probability that $A$ occurs, multiplied by the conditional probability that $B$ then occurs, calculated with the assumption that $A$ occurred. Thus, if $A$ is the probability of drawing a heart on the first draw, the probability of $A$ is $13 / 52$. The probability of drawing a heart on the second draw, given that the first draw yielded a heart, is $12 / 51$ since there remain 12 hearts in the deck. The probability of drawing two hearts is then $(13 / 52)(12 / 51)=1 / 17$, as found previously.
In quantum mechanics we must deal with probabilities involving a continuous variable, for example, the $x$ coordinate. It does not make much sense to talk about the probability of a particle being found at a particular point such as $x=0.5000 \ldots$, since there are an infinite number of points on the $x$ axis, and for any finite number of measurements we make, the probability of getting exactly $0.5000 \ldots$ is vanishingly small. Instead we talk of the probability of finding the particle in a tiny interval of the $x$ axis lying between $x$ and $x+d x, d x$ being an infinitesimal element of length. This probability will naturally be proportional to the length of the interval, $d x$, and will vary for different regions of the $x$ axis. Hence the probability that the particle will be found between $x$ and $x+d x$ is equal to $g(x) d x$, where $g(x)$ is some function that tells how the probability varies over the $x$ axis. The function $g(x)$ is called the probability density, since it is a probability per unit length. Since probabilities are real, nonnegative numbers, $g(x)$ must be a real function that is everywhere nonnegative. The wave function $\Psi$ can take on negative and complex values and is not a probability density. Quantum mechanics postulates that the probability density is $|\Psi|^{2}$ [Eq. (1.15)].
What is the probability that the particle lies in some finite region of space $a \leq x \leq b$ ? To find this probability, we sum up the probabilities $|\Psi|^{2} d x$ of finding the particle in all
the infinitesimal regions lying between $a$ and $b$. This is just the definition of the definite integral
\(
\begin{equation*}
\int_{a}^{b}|\Psi|^{2} d x=\operatorname{Pr}(a \leq x \leq b) \tag{1.23}
\end{equation*}
\)
where Pr denotes a probability. A probability of 1 represents certainty. Since it is certain that the particle is somewhere on the $x$ axis, we have the requirement
\(
\begin{equation*}
\int_{-\infty}^{\infty}|\Psi|^{2} d x=1 \tag{1.24}
\end{equation*}
\)
When $\Psi$ satisfies (1.24), it is said to be normalized. For a stationary state, $|\Psi|^{2}=|\psi|^{2}$ and $\int_{-\infty}^{\infty}|\psi|^{2} d x=1$.
## EXAMPLE
A one-particle, one-dimensional system has $\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \mathrm{~nm}$. At $t=0$, the particle's position is measured. (a) Find the probability that the measured value lies between $x=1.5000 \mathrm{~nm}$ and $x=1.5001 \mathrm{~nm}$. (b) Find the probability that the measured value is between $x=0$ and $x=2 \mathrm{~nm}$. (c) Verify that $\Psi$ is normalized.
(a) In this tiny interval, $x$ changes by only 0.0001 nm , and $\Psi$ goes from $e^{-1.5000} \mathrm{~nm}^{-1 / 2}=0.22313 \mathrm{~nm}^{-1 / 2}$ to $e^{-1.5001} \mathrm{~nm}^{-1 / 2}=0.22311 \mathrm{~nm}^{-1 / 2}$, so $\Psi$ is nearly constant in this interval, and it is a very good approximation to consider this interval as infinitesimal. The desired probability is given by (1.15) as
\(
\begin{aligned}
|\Psi|^{2} d x=a^{-1} e^{-2|x| / a} d x & =(1 \mathrm{~nm})^{-1} e^{-2(1.5 \mathrm{~nm}) /(1 \mathrm{~nm})}(0.0001 \mathrm{~nm}) \\
& =4.979 \times 10^{-6}
\end{aligned}
\)
(See also Prob. 1.14.)
(b) Use of Eq. (1.23) and $|x|=x$ for $x \geq 0$ gives
\(
\begin{aligned}
\operatorname{Pr}(0 \leq x \leq 2 \mathrm{~nm}) & =\int_{0}^{2 \mathrm{~nm}}|\Psi|^{2} d x=a^{-1} \int_{0}^{2 \mathrm{~nm}} e^{-2 x / a} d x \\
& =-\left.\frac{1}{2} e^{-2 x / a}\right|_{0} ^{2 \mathrm{~nm}}=-\frac{1}{2}\left(e^{-4}-1\right)=0.4908
\end{aligned}
\)
(c) Use of $\int_{-\infty}^{\infty} f(x) d x=\int_{-\infty}^{0} f(x) d x+\int_{0}^{\infty} f(x) d x,|x|=-x$ for $x \leq 0$, and $|x|=x$ for $x \geq 0$, gives
\(
\begin{aligned}
\int_{-\infty}^{\infty}|\Psi|^{2} d x & =a^{-1} \int_{-\infty}^{0} e^{2 x / a} d x+a^{-1} \int_{0}^{\infty} e^{-2 x / a} d x \\
& =a^{-1}\left(\left.\frac{1}{2} a e^{2 x / a}\right|_{-\infty} ^{0}\right)+a^{-1}\left(-\left.\frac{1}{2} a e^{-2 x / a}\right|_{0} ^{\infty}\right)=\frac{1}{2}+\frac{1}{2}=1
\end{aligned}
\)
EXERCISE For a system whose state function at the time of a position measurement is $\Psi=\left(32 a^{3} / \pi\right)^{1 / 4} x e^{-a x^{2}}$, where $a=1.0000 \mathrm{~nm}^{-2}$, find the probability that the particle is found between $x=1.2000 \mathrm{~nm}$ and 1.2001 nm . Treat the interval as infinitesimal.
(Answer: 0.0000258.)
There has been much controversy about the proper definition of probability. One definition is the following: If an experiment has $n$ equally probable outcomes, $m$ of which are favorable to the occurrence of a certain event $A$, then the probability that $A$ occurs is $m / n$. Note that this definition is circular, since it specifies equally probable outcomes when probability is what we are trying to define. It is simply assumed that we can recognize equally probable outcomes. An alternative definition is based on actually performing the experiment many times. Suppose that we perform the experiment $N$ times and that in $M$ of these trials the event $A$ occurs. The probability of $A$ occurring is then defined as
\(
\lim _{N \rightarrow \infty} \frac{M}{N}
\)
Thus, if we toss a coin repeatedly, the fraction of heads will approach $1 / 2$ as we increase the number of tosses.
For example, suppose we ask for the probability of drawing a heart when a card is picked at random from a standard 52 -card deck containing 13 hearts. There are 52 cards and hence 52 equally probable outcomes. There are 13 hearts and hence 13 favorable outcomes. Therefore, $m / n=13 / 52=1 / 4$. The probability for drawing a heart is $1 / 4$.
Sometimes we ask for the probability of two related events both occurring. For example, we may ask for the probability of drawing two hearts from a 52 -card deck, assuming we do not replace the first card after it is drawn. There are 52 possible outcomes of the first draw, and for each of these possibilities there are 51 possible second draws. We have $52 \cdot 51$ possible outcomes. Since there are 13 hearts, there are $13 \cdot 12$ different ways to draw two hearts. The desired probability is $(13 \cdot 12) /(52 \cdot 51)=1 / 17$. This calculation illustrates the theorem: The probability that two events $A$ and $B$ both occur is the probability that $A$ occurs, multiplied by the conditional probability that $B$ then occurs, calculated with the assumption that $A$ occurred. Thus, if $A$ is the probability of drawing a heart on the first draw, the probability of $A$ is $13 / 52$. The probability of drawing a heart on the second draw, given that the first draw yielded a heart, is $12 / 51$ since there remain 12 hearts in the deck. The probability of drawing two hearts is then $(13 / 52)(12 / 51)=1 / 17$, as found previously.
In quantum mechanics we must deal with probabilities involving a continuous variable, for example, the $x$ coordinate. It does not make much sense to talk about the probability of a particle being found at a particular point such as $x=0.5000 \ldots$, since there are an infinite number of points on the $x$ axis, and for any finite number of measurements we make, the probability of getting exactly $0.5000 \ldots$ is vanishingly small. Instead we talk of the probability of finding the particle in a tiny interval of the $x$ axis lying between $x$ and $x+d x, d x$ being an infinitesimal element of length. This probability will naturally be proportional to the length of the interval, $d x$, and will vary for different regions of the $x$ axis. Hence the probability that the particle will be found between $x$ and $x+d x$ is equal to $g(x) d x$, where $g(x)$ is some function that tells how the probability varies over the $x$ axis. The function $g(x)$ is called the probability density, since it is a probability per unit length. Since probabilities are real, nonnegative numbers, $g(x)$ must be a real function that is everywhere nonnegative. The wave function $\Psi$ can take on negative and complex values and is not a probability density. Quantum mechanics postulates that the probability density is $|\Psi|^{2}$ [Eq. (1.15)].
What is the probability that the particle lies in some finite region of space $a \leq x \leq b$ ? To find this probability, we sum up the probabilities $|\Psi|^{2} d x$ of finding the particle in all
the infinitesimal regions lying between $a$ and $b$. This is just the definition of the definite integral
\(
\begin{equation*}
\int_{a}^{b}|\Psi|^{2} d x=\operatorname{Pr}(a \leq x \leq b) \tag{1.23}
\end{equation*}
\)
where Pr denotes a probability. A probability of 1 represents certainty. Since it is certain that the particle is somewhere on the $x$ axis, we have the requirement
\(
\begin{equation*}
\int_{-\infty}^{\infty}|\Psi|^{2} d x=1 \tag{1.24}
\end{equation*}
\)
When $\Psi$ satisfies (1.24), it is said to be normalized. For a stationary state, $|\Psi|^{2}=|\psi|^{2}$ and $\int_{-\infty}^{\infty}|\psi|^{2} d x=1$.
## EXAMPLE
A one-particle, one-dimensional system has $\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \mathrm{~nm}$. At $t=0$, the particle's position is measured. (a) Find the probability that the measured value lies between $x=1.5000 \mathrm{~nm}$ and $x=1.5001 \mathrm{~nm}$. (b) Find the probability that the measured value is between $x=0$ and $x=2 \mathrm{~nm}$. (c) Verify that $\Psi$ is normalized.
(a) In this tiny interval, $x$ changes by only 0.0001 nm , and $\Psi$ goes from $e^{-1.5000} \mathrm{~nm}^{-1 / 2}=0.22313 \mathrm{~nm}^{-1 / 2}$ to $e^{-1.5001} \mathrm{~nm}^{-1 / 2}=0.22311 \mathrm{~nm}^{-1 / 2}$, so $\Psi$ is nearly constant in this interval, and it is a very good approximation to consider this interval as infinitesimal. The desired probability is given by (1.15) as
\(
\begin{aligned}
|\Psi|^{2} d x=a^{-1} e^{-2|x| / a} d x & =(1 \mathrm{~nm})^{-1} e^{-2(1.5 \mathrm{~nm}) /(1 \mathrm{~nm})}(0.0001 \mathrm{~nm}) \\
& =4.979 \times 10^{-6}
\end{aligned}
\)
(See also Prob. 1.14.)
(b) Use of Eq. (1.23) and $|x|=x$ for $x \geq 0$ gives
\(
\begin{aligned}
\operatorname{Pr}(0 \leq x \leq 2 \mathrm{~nm}) & =\int_{0}^{2 \mathrm{~nm}}|\Psi|^{2} d x=a^{-1} \int_{0}^{2 \mathrm{~nm}} e^{-2 x / a} d x \\
& =-\left.\frac{1}{2} e^{-2 x / a}\right|_{0} ^{2 \mathrm{~nm}}=-\frac{1}{2}\left(e^{-4}-1\right)=0.4908
\end{aligned}
\)
(c) Use of $\int_{-\infty}^{\infty} f(x) d x=\int_{-\infty}^{0} f(x) d x+\int_{0}^{\infty} f(x) d x,|x|=-x$ for $x \leq 0$, and $|x|=x$ for $x \geq 0$, gives
\(
\begin{aligned}
\int_{-\infty}^{\infty}|\Psi|^{2} d x & =a^{-1} \int_{-\infty}^{0} e^{2 x / a} d x+a^{-1} \int_{0}^{\infty} e^{-2 x / a} d x \\
& =a^{-1}\left(\left.\frac{1}{2} a e^{2 x / a}\right|_{-\infty} ^{0}\right)+a^{-1}\left(-\left.\frac{1}{2} a e^{-2 x / a}\right|_{0} ^{\infty}\right)=\frac{1}{2}+\frac{1}{2}=1
\end{aligned}
\)
EXERCISE For a system whose state function at the time of a position measurement is $\Psi=\left(32 a^{3} / \pi\right)^{1 / 4} x e^{-a x^{2}}$, where $a=1.0000 \mathrm{~nm}^{-2}$, find the probability that the particle is found between $x=1.2000 \mathrm{~nm}$ and 1.2001 nm . Treat the interval as infinitesimal.
(Answer: 0.0000258.)