Superposition of Wavefunctions

Quantum mechanics is a linear theory: if $\psi_1(x)$ and $\psi_2(x)$ are both valid wavefunctions, then any linear combination

$\Psi(x) = c_1\,\psi_1(x) + c_2\,\psi_2(x)$

is also a valid wavefunction — provided it is normalised. This is the superposition principle.

Why normalisation matters

The Born rule assigns a probability density to $\Psi$ :

$p(x) = |\Psi(x)|^2.$

For this to be a genuine probability density it must integrate to 1 over all space:

$\int_{-\infty}^{\infty} |\Psi(x)|^2\,dx = 1.$

If you build a superposition from individually normalised, mutually orthogonal states $\psi_1$ and $\psi_2$ ,

$\int \psi_1^*(x)\,\psi_1(x)\,dx = 1, \quad \int \psi_2^*(x)\,\psi_2(x)\,dx = 1, \quad \int \psi_1^*(x)\,\psi_2(x)\,dx = 0,$

and you form $\Psi = A(\psi_1 + \psi_2)$ with a real constant $A$ , then

$\int |\Psi|^2\,dx = |A|^2\left(\int|\psi_1|^2\,dx + \int\psi_1^*\psi_2\,dx + \int\psi_2^*\psi_1\,dx + \int|\psi_2|^2\,dx\right) = |A|^2(1 + 0 + 0 + 1) = 2|A|^2.$

Setting $2|A|^2 = 1$ gives

$A = \frac{1}{\sqrt{2}}.$

The cross terms vanished because of orthogonality. This is always the key step: orthogonal components contribute independently to the total probability.

The amplitudes carry the physics

Once normalised, the coefficients encode how much of each state is present. For $\Psi = \tfrac{1}{\sqrt{2}}(\psi_1 + \psi_2)$ , the probability of finding the particle in energy eigenstate $\psi_n$ (if you measure energy) is $|c_n|^2$ :

$P(\psi_1) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}, \qquad P(\psi_2) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}.$

Equal weights give equal probabilities — a 50/50 quantum lottery.

General case

For a superposition of $n$ orthonormal states with equal amplitudes,

$\Psi = A\sum_{k=1}^{n}\psi_k,$

the same calculation gives $n|A|^2 = 1$ , hence $A = 1/\sqrt{n}$ . Unequal weights $c_k$ are handled the same way: compute $\sum_k|c_k|^2$ from the orthonormality relations and divide each coefficient by the square root of that sum.

Try it

This is a numerical exercise — return a number. Two individually normalised, orthogonal wavefunctions are combined with equal weight: $\Psi = A(\psi_1 + \psi_2)$ . What positive real constant $A$ makes $\Psi$ normalised?

Run your code to see the quantum state.