Problem Set 1 - Siavash Davani's Website

Consider the function given below

\rho(x) = A \, e^{-(ax^2+bx+c)}, \quad\quad a > 0

(1)

What value should $A$ have as a function of $a$ , $b$ , and $c$ , for $\rho(x)$ to be a probability distribution?
Solution
For $\rho(x)$ to be a probability distribution, we must have $\int_{-\infty}^{+\infty} \rho(x) \, dx = 1$ which means
$\begin{align} \int_{-\infty}^{+\infty} A \, e^{-(ax^2+bx+c)} \, dx &= A \, e^{-c} \int_{-\infty}^{+\infty} e^{-(ax^2+bx)} \, dx \\ &= A \, e^{-c} \, e^{\frac{b^2}{4a}} \int_{-\infty}^{+\infty} e^{-(ax^2+bx+\frac{b^2}{4a})} \, dx \\ &= A \, e^{-c} \, e^{\frac{b^2}{4a}} \int_{-\infty}^{+\infty} e^{-a(x+\frac{b}{2a})^2} \, dx \end{align}$
(2)
If we change variable using $x' = \sqrt{a} \left(x+\frac{b}{2a}\right)$ , we have $dx'=\sqrt{a} \, dx$ , and this means the integral will become
$\begin{align} \int_{-\infty}^{+\infty} A \, e^{-(ax^2+bx+c)} \, dx &= A \, e^{-c} \, e^{\frac{b^2}{4a}} \, \frac{1}{\sqrt{a}} \int_{-\infty}^{+\infty} e^{-x'^2} \, dx' \\ &= A \, e^{-c} \, e^{\frac{b^2}{4a}} \, \frac{1}{\sqrt{a}} \, \sqrt{\pi}, \end{align}$
(3)
where in the last step we used the well known Gaussian integral $\int_{-\infty}^{+\infty} e^{-x^2} \, dx = \sqrt{\pi}$ .
This means $A$ must be
$A = \sqrt{\frac{a}{\pi}} \, e^{(c-\frac{b^2}{4a})}.$
(4)
Calculate the expectation value $\langle x \rangle$ using this probability distribution.
Solution
We have
$\langle x \rangle = \int_{-\infty}^{+\infty} A \, x \, e^{-(ax^2+bx+c)} \, dx.$
(5)
There are many ways to calculate this integral. You can as well look up the properties of Gaussian distributions. But let us see one way to solve this in the following. In the previous task, we derived
$I(A, a, b, c) = \int_{-\infty}^{+\infty} A \, e^{-(ax^2+bx+c)} \, dx = A \, e^{-c} \, e^{\frac{b^2}{4a}} \, \frac{1}{\sqrt{a}} \, \sqrt{\pi}.$
(6)
If we pay attention closely, we can see that if we take the derivative with respect to the variable $b$ in Eq 6, we have
$\pdv{I}{b} = \int_{-\infty}^{+\infty} -A \, x \, e^{-(ax^2+bx+c)} \, dx = \left( \frac{b}{2a} \right) A \, e^{-c} \, e^{\frac{b^2}{4a}} \, \frac{1}{\sqrt{a}} \, \sqrt{\pi}$
(7)
This means, the expectation value is
$\langle x \rangle = -\pdv{I}{b} = -\frac{b}{2a},$
(8)
by putting $A$ from Eq. 4 in the derivative.
Calculate $\langle x^2 \rangle$ and the standard deviation σ.
Solution
We can see that similar to what we did for calculating $\langle x \rangle$ , we can derive
$\begin{align} \langle x^2 \rangle &= \int_{-\infty}^{+\infty} A \, x^2 \, e^{-(ax^2+bx+c)} \, dx = -\pdv{I}{a} \\ &= - \left( -\frac{b^2}{4a^2}e^{\frac{b^2}{4a}} \, \frac{1}{\sqrt{a}} - e^{\frac{b^2}{4a}} \, \frac{1}{2a\sqrt{a}} \right) A \, e^{-c} \, \sqrt{\pi} \\ &= \frac{b^2}{4a^2} + \frac{1}{2a}. \end{align}$
(9)
And for the standard deviation, we have
$\begin{align} \sigma^2 &= \langle x^2 \rangle - \langle x \rangle^2 \\ &= \frac{b^2}{4a^2} + \frac{1}{2a} - \frac{b^2}{4a^2} = \frac{1}{2a}. \end{align}$
(10)
Therefore
$\sigma = \sqrt{\frac{1}{2a}}.$
(11)

Exercise 2 (Normalization)

Assume the wavefunction corresponding to a particle is given as

\psi(x) = \begin{cases} 0 & x < -1 \\ A \, ( x + 1 ) & -1 \le x < 0 \\ A \, e^{-\frac{x}{2}} & x \ge 0 \end{cases}

(12)

Find $A$ such that $\psi(x)$ is a valid normalized wavefunction.
Solution
The point you have to be careful about is that in quantum mechanics, the square of the wavefunction $\abs{\psi(x)}^2$ corresponds to a probability distribution not the wavefunction $\psi(x)$ itself. Therefore, in this task you should calculate
$I_1 = \int_{-1}^{0} \abs{\psi(x)}^2 \, dx = \frac{1}{3} A^2,$
(13)
and
$I_2 = \int_{0}^{\infty} \abs{\psi(x)}^2 \, dx = A^2,$
(14)
which leads to
$\int_{-\infty}^{\infty} \abs{\psi(x)}^2 = \frac{4}{3} A^2 = 1,$
(15)
for normalization. Therefore,
$A = \frac{\sqrt{3}}{2}.$
(16)
Plot the normalized wavefunction $\psi(x)$ .
Plot the corresponding probability distribution $\abs{\psi(x)}^2$ .
Which point $x$ is the most probable place to find the particle if we measure its position?
Solution
The highest value of the probability is at the location $x=0$ and the probability density is $P(0)=\abs{\psi(0)}^2=\frac{3}{4}$ there.
What will be the average value of the position? Does it coincide with the most probable position?
Solution
We must calculate
$I_3 = \int_{-1}^{0} x \, \abs{\psi(x)}^2 \, dx = -\frac{1}{12} A^2 = -\frac{1}{16},$
(17)
and
$I_4 = \int_{0}^{\infty} x \, \abs{\psi(x)}^2 \, dx = A^2 = \frac{3}{4}.$
(18)
Therefore, we have
$\begin{align} \langle x \rangle &= \int_{-\infty}^{\infty} x \, \abs{\psi(x)}^2 \, dx \\ &= I_3 + I_4 = \frac{11}{16}. \end{align}$
(19)
As we can see, the most probable position of the particle is not necessarily equal to the expected value of the position (which corresponds to the statistical average of repeating the measurement of the position on many identical copies of the quantum mechanical particle).