# Conflicting dimensions of wave function

In my textbook, there is an equation expressing a nonperiodic wave function as an infinite sum of waves over a continuum of wave numbers $$\psi(x) = \int_{-\infty}^\infty A(k)e^{ikx}dk$$

I know that $$\mathrm{probability \ density} = |\psi|^2 \\ k=\frac{2\pi}{\lambda}$$

I determine the unit of $$\psi(x)$$ to be $$m^{-1/2}$$ by assuming that the probability density has units of $$m^{-1}$$ in 1D space. Since a single plane wave function is given by $$\psi(x) = Ae^{ikx}$$, by taking the exponent to be unitless, I get the unit of $$A$$ to be $$m^{-1/2}$$.

However, when I look at the integral expression, $$\psi(x)$$ appears to be the area under the curve $$f(k) = A(k)e^{ikx}$$. This means the unit of $$\psi(x)$$ would become $$m^{-1}m^{-1/2}=m^{-3/2}$$ considering that the unit of $$k$$ is $$m^{-1}$$ and the unit of $$A(k)$$ remains unchanged.

The dimensions do not match in this analysis. What am I missing?

Edit: Thank everyone for the answers. Multiple people have pointed out that the units of $$A(k)$$ and $$A$$ are not the same, but the textbook states the following:

The amplitude $$A(k)$$ of the plane wave is naturally a function of k, for it tells us how much of each different wave number goes into the sum.

I assume it implies that $$A(k)$$ is the amplitude of a single constituent plane wave of wave number $$k$$? If not, why is $$A(k)$$ expressed as "amplitude" when it is really amplitude per wavenumber?

• You have provided two definitions of $\psi$. The A(k) in the integral does not have the same units as A in your single plane wave equation.
– Ali
Nov 16, 2020 at 6:32

I think this is a proper question and does require an explanation. Because similar notation (see, for example, the hyperphysics page) is used in many places which can be confusing for anyone, especially for beginner students.

Let me try to answer your question with an analogy. As you probably know, work in physics is defined as scalar product of force and displacement as $$W=\vec{F}\cdot \Delta \vec{x}$$ One can use this formula as long as the force is not a function of position, that is, constant over the whole displacement. Otherwise it should be calculated as an integral $$W=\int{\vec{F}\cdot d\vec{x}}$$ Now in your case, the wave function in one dimension is defined as a plane wave multiplied by a constant amplitude as $$\psi\left(x \right )=Ae^{ikx}$$ If, however, one wants to define a wave function as a superposition of multiple plane waves one can write, for the discrete case $$\psi\left(x \right )=\sum_{j}A_je^{ikx}$$ which for the continuous case should be written as $$\psi\left(x \right )=\int e^{ikx}dA$$ Now the trick here is that both $$e^{ikx}$$ and $$A$$ are functions of $$k$$ therefore one can write $$\psi\left(x \right )=\int \frac{\partial A}{\partial k}e^{ikx}dk$$ which is then (probably) written as $$\psi\left(x \right )=\int A\left(k \right) e^{ikx}dk$$ This answers your question, that is, $$A$$ and $$A\left(k \right)$$ are not the same thing and written in this way the units are correct.

In $$n-$$dimensions of space, the dimension of the wavefunction is is always $$\text{Dim}[\psi] = -\frac{n}{2}.$$ This can be seen very easily since we know that the probability is dimensionless, we must have that:

$$\int_V |\psi|^2 \text{d}V = 1.$$

The dimensions of $$\text{d}V$$ are the same as that of volume in $$n-$$dimensions, and so $$\text{Dim}[\text{d}V] = L^n$$, and therefore $$\text{Dim}[|\psi|^2] = L^{-n}$$, for the right hand side to be dimensionless, from which the above result follows simply. In one dimension, the dimension of $$\psi(x)$$ is therefore $$L^{-1/2}$$.

As to your example, as @Ali points out, you are trying to compare the dimensions of

$$\int_{-\infty}^\infty A(k) e^{ikx}\text{d}k \quad \quad \text{and}\quad\quad A e^{ikx}.$$

Clearly, even though you have given them the same name, $$A(k)$$ and $$A$$ cannot have the same dimension, since $$\text{d}k$$ is not dimensionless.

Your error comes when you write $$\psi(x)=Ae^{ikx}$$. This directly conflicts with your original statement, $$\psi(x)=\int Ae^{ikx}dk.$$ Noting that $$|\psi|^2dx$$ should be a unitless quantity as you have done, indeed $$\psi$$ should have dimension $$m^{-1/2}$$. Since $$k$$ has units $$m^{-1}$$ (which can be deduced from $$ikx$$ being unitless, regardless of this additional statement $$k=2\pi/\lambda$$), it follows that $$Adk$$ should have units $$m^{-1/2}$$, and hence $$A$$ must have units $$m^{1/2}$$.