Exponential Wavefunction for Infinite Potential Well Problem

[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.

 

[PeroK]

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.

 

[a1234]

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?

 

[PeroK]

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.

 

[a1234]

I see, that makes sense. Thank you for the help!

 

[etotheipi]

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?

 

[PeroK]

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.

 

[vanhees71]

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.