To prove that the wave functions for a particle in a one-dimensional box are orthogonal, we need to show that their inner product (also known as the overlap integral) is equal to zero when the wave functions correspond to different energy eigenstates.
Let's consider a particle in a one-dimensional box of length L. The wave function ψ_n(x) represents the nth energy eigenstate, where n is a positive integer. These wave functions can be expressed as:
ψ_n(x) = sqrt(2/L) * sin(nπx/L),
where sqrt represents the square root, and sin denotes the sine function.
To prove orthogonality, we need to evaluate the integral of the product of two different wave functions over the entire length of the box. Let's consider two different energy eigenstates, m and n, where m ≠ n.
∫ ψ_m*(x) ψ_n(x) dx = ∫ (sqrt(2/L) * sin(mπx/L)) * (sqrt(2/L) * sin(nπx/L)) dx,
Using the trigonometric identity sin(a) sin(b) = (1/2)[cos(a-b) - cos(a+b)], we can simplify the integral:
∫ ψ_m*(x) ψ_n(x) dx = (2/L) ∫ sin(mπx/L) sin(nπx/L) dx = (1/L) ∫ [cos((m-n)πx/L) - cos((m+n)πx/L)] dx.
Now, let's evaluate each term of the integral separately.
For the first term, cos((m-n)πx/L), when integrated over the entire length of the box, the result is zero since the cosine function has a period of 2π and the argument (m-n)πx/L ranges from 0 to L.
For the second term, cos((m+n)πx/L), when integrated over the entire length of the box, the result is zero for the same reason as above.
Therefore, both terms of the integral evaluate to zero, and we have:
∫ ψ_m*(x) ψ_n(x) dx = 0.
This demonstrates that the wave functions ψ_m(x) and ψ_n(x) are orthogonal when m ≠ n.
Hence, the wave functions for a particle in a one-dimensional box are orthogonal.