Probability function area vs Sum of Probabilities

Question

I've a small confusion about Probabilities in Quantum Mechanics.

If a particle can move only on the positive $x-axis$ between $x=1$ and $x=\infty$, then its probability function must satisfy $\int_1^{\infty}f(x)dx=1$, right?

$f(x)=\frac{1}{x}$ is one such function. So, I think it's possible that the probability of finding a particle whose motion is restricted between $x=1$ and $x=\infty$ is described by this function.

Now, if we use this function, then the probability of finding the particle at $x=2$ is $\frac{1}{2}$, then probability of finding it at $x=3$ is $\frac{1}{3}$. The probability of finding it at $x=4$ is $\frac{1}{4}$.

So, the probability of finding the particle either at $x=1$, $x=2$ or $x=3$ is:

$$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=1.083333333333.....>1$$

How can it be greater than 1? Also, the probability of finding the particle at $x=1$ will be 1!. How is that possible?

Answer 1

The error is in the statement "the probability of finding the particle at $x$". This probability is actually $0$. The function $f$ of your question is a probability density. The product of this $f$ multiplied by a small interval $dx$, i.e. $f(x_0)dx$, gives you the probability of finding the particle in a bin of width $dx$ centered at $x_0$.

The correct understanding that $f(x)$ is a probability density, and recognizing that the probability at one specific point is $0$ is essential and a typical trick question on quantum mechanics exams. The distinction is clear if you realize that, given $dx$ has unit of length and that $\int f(x)dx$ must be dimensionless since it's a probability, it must be that $f(x)$ has unit of inverse length. Thus $f(x)$ at $x=1$ is equal to $1/m$ (i.e. $1$ per meter), not $1$.

In addition, it is the integral of $f(x)$ that must be one, not the function itself. This does not prevent $f(x)$ itself from being greater than $1$ since what matters is the area under the curve. In your setup, consider the function $$ \bar{f}(x)=\left\{\begin{array}{cc} 2&\hbox{if } 1<x<3/2\, ,\\ 0&\hbox{elsewhere}.\end{array}\right. $$ This function takes the value $2$ over an interval of width $1/2$ and is $0$ elsewhere, so the area under the curve of this function is $1$, even if the function itself is greater than $1$ over a finite interval. Of course it makes no sense to suggest that the probability at $x=1.25$ is $2$ simply because $\bar{f}(1.25)=2$.

As an analogy imagine a glass of water. Asking how much water (in kg) is in a small volume $dV$ makes sense: this amount is just $\rho dV$, where $\rho$ is the mass density of the water. Asking how much water (in kg) is at one point makes no sense - it is certainly not related to the density of water.

Answer 2

$\int_1^b \frac{1}{x} dx=\ln(b)$ which does not equal 1, it diverges as $b \to \infty$.

Furthermore, the probability density of finding the particle at $a$ is $\frac{1}{x}$. So you cannot add the probability densities at $1,2,3,4...$ and expect to get 1. This sum (the harmonic series) also diverges - ie has no finite limiting value.

Answer 3

1. "If a particle can move only on the positive x−axis between $x=1$ and $x=∞$, then its probability function must satisfy $\int_{1}^{\infty} f(x)dx=1$, right?"

It Should be "If a particle can move only on the positive x−axis between $x=1$ and $x=∞$, then its probability density function $f(x)$ must satisfy $\int_{1}^{\infty} f(x)dx=1$"

2. Probability can't be over 1, but not the case for probability density that can be over 1(must be positive). You pick $f(x)=\frac{1}{x}$ as the probability density function, however, it is not a real or meanningful one since it diverges as pointed by sammy.

3. Once you know the probability density function $f(x)$, then via the normalization formula $\int_{-\infty}^{\infty} f(x)dx=1$ to normalize it. The probability in the region $a<x<b$ can be obtained with $P(a<x<b)=\int_{a}^{b} f(x)dx$