# Exponential Wavefunction for Infinite Potential Well Problem

• a1234
In summary, the conversation discusses the use of boundary conditions to find a constant in the normalization integral for a wave function. It is mentioned that the constant can be a complex number, but it is implied that it should have a modulus of 1 to maintain normalization. It is also noted that different values for this constant can lead to mathematically and physically equivalent solutions. The importance of this concept is highlighted in relation to half-integer eigen values for angular momentum and the representation of fermions in nature.

#### a1234

Homework Statement
I am asked to solve the Schrodinger Equation for the "particle in a box" problem (with box width going from 0 to a), but using complex exponentials instead of sin and cos.
Relevant Equations
sin(x) = (exp(ix) - exp(-ix))/(2i);
psi(x) = A(exp(ikx) - exp(ikx))
psi(x) = A*2i*sin(kx)
psi(x) = A*sin(kx)
Using the boundary conditions where psi is 0, I found that k = n*pi/a, since sin(x) is zero when k*a = 0.

I set up my normalization integral as follows:

A^2 * integral from 0 to a of (((exp(ikx) - exp(-ikx))*(exp(-ikx) - exp(ikx)) dx) = 1

After simplifying, and accounting for the fact that sin(2*n*pi) = 0, I arrived at the following expression:

2*a*A^2 = 1
A = 1/√(2a)

However, once we substitute the above expression for A in the original wave function, we get psi(x) = i√(2/a)*sin(kx).

I would like to know if there is a way to avoid the extra factor of i in the wavefunction. I've read online that the extra i is often absorbed into the constant A, but is there a way around this? I've been told by my instructor that the constant A is supposed to be complex, and that the final wavefunction is supposed to have real components.

That's a perfectly legitimate form of the solution. It's purely a matter of convention and convenience to choose a real normalisation constant. If you need an arbitrary complex number of unit modulus, then ##i## is just as valid as ##1##. As is any other complex number of unit modulus.

• vanhees71, a1234, etotheipi and 1 other person
Ok. Is there any way of demonstrating that the solutions are equivalent? How does having a complex number in the constant not affect the significance of the wavefunction?

a1234 said:
Ok. Is there any way of demonstrating that the solutions are equivalent? How does having a complex number in the constant not affect the significance of the wavefunction?
You're looking for solutions to the SDE, which is linear. If ##\Psi## is a solution, then so is ##\alpha \Psi## for any complex number ##\alpha## - which is often called a phase factor.

It's a slightly subtler point that the solutions are physically equivalent. To see this, you could note that the expectation value of any observable is not changed by a phase factor.

In any case, the solutions are mathematically and physically equivalent.

• etotheipi and a1234
I see, that makes sense. Thank you for the help!

PeroK said:
You're looking for solutions to the SDE, which is linear. If Ψ is a solution, then so is αΨ for any complex number α - which is often called a phase factor.

It's a slightly subtler point that the solutions are physically equivalent. To see this, you could note that the expectation value of any observable is not changed by a phase factor.

Although ##\alpha \psi## is a solution for any ##\alpha \in \mathbb{C}##, I suppose it's implied that in Physics that number has to satisfy ##|\alpha| = 1 \iff \alpha = e^{i \delta}##, otherwise ##\int \alpha^*\psi^*(x) \alpha \psi(x) dx## is no longer normalised?

etotheipi said:
Although ##\alpha \psi## is a solution for any ##\alpha \in \mathbb{C}##, I suppose it's implied that in Physics that number has to satisfy ##|\alpha| = 1 \iff \alpha = e^{i \delta}##, otherwise ##\int \alpha^*\psi^*(x) \alpha \psi(x) dx## is no longer normalised?
Yes, if ##\psi## is already normalised, then we want to keep it that way.

• etotheipi
That's a very important point! In quantum mechanics pure states are not represented by the Hilbert-space vectors ("kets"), ##|\psi \rangle## but rather by the statistical operator ##\hat{\rho}=|\psi \rangle \langle \psi|## (with ##|\psi \rangle## understood to be normalized, ##\langle \psi|\psi \rangle=1##). This shows that two state vectors ##|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle## with ##\varphi \in \mathbb{R}## lead to the same state ##\hat{\rho}'=\hat{\rho}=|\psi \rangle \langle \psi|##.

That's very important. E.g., because of this, it makes sense to have half-integer eigen values for angular momentum. Since by definition angular momentum are the generators of rotation, all the half-integer values ##S## following from the angular-momentum commutation relations alone, can provide a valid description of nature, although a rotation by ##2 \pi## changes the wave function by a factor ##-1##. If the wave function itself were representing measurable states, this would make no sense, but as we well know, all the matter around us consists of spin-1/2 particles being fermions.

• etotheipi