# Illustrating the correspondence between 1-forms and vectors using de Broglie waves

I was asked by a student to clarify the issues surrounding an exercise in the famous book Gravitation written by Misner, Thorne and Wheeler (MTW). The exercise appears as follows in Chapter 2, Section 5:

The key point of this section is that Equation (2.14), the defining equation of 1-forms, can be shown to be physically valid (as well as being a just a mathematical definition) using de Broglie waves in quantum mechanics. The notation in MTW is not ideal, so we will replace the notation $\langle \mathbf{\tilde{p}}, \mathbf{v} \rangle$ for a 1-form evaluated at a vector $\mathbf{v}$ by the notation $\mathbf{\tilde{p}}(\mathbf{v})$. What MTW are then saying is that given any vector $\mathbf{p}$ we can define a corresponding 1-form as

$\mathbf{\tilde{p}} = \langle \mathbf{p}, \ \rangle$

which is to be viewed as a function waiting for a vector input (to be placed in the empty space on the right-hand side of the angle brackets). When the vector input $\mathbf{v}$ is supplied, the 1-form will then yield the number

$\mathbf{\tilde{p}}(\mathbf{v}) = \langle \mathbf{p}, \mathbf{v} \rangle = \mathbf{p} \cdot \mathbf{v}$

In Exercise 2.1 we are asked to verify the validity of this equation using the de Broglie wave

$\psi = e^{i \phi} = exp[i(\mathbf{k}\cdot \mathbf{x}- \omega t)]$

The phase is the angular argument $\phi = \mathbf{k}\cdot \mathbf{x} - \omega t$ which specifies the position of the wave from some starting point. The phase is parameterised by the wave vector $\mathbf{k}$ which is such that $|\mathbf{k}| = 2 \pi/\lambda$ where $\lambda$ is the wavelength, and by the angular frequency $\omega = 2 \pi f$ where $f$ is the frequency of the relevant oscillator.

It is a well known fact (and it is easy to verify) that given any real-valued function of a vector $\phi(\mathbf{x})$, the gradient vector $\partial \phi/\partial \mathbf{x}$ is orthogonal to the level surfaces of $\phi$. In the case of the phase of a de Broglie wave we have

$\frac{\partial \phi}{\partial \mathbf{x}} = \mathbf{k}$

so the wave vector is the (position) gradient vector of the phase $\phi$ and therefore $\mathbf{k}$ must be orthogonal to loci of constant phase.

In the case of circular waves, for example, these loci of constant phase are circles with centre at the source of the waves and the wave vectors $\mathbf{k}$ point radially outwards at right angles to them, as indicated in the diagram.

To get a diagrammatic understanding of the relationship between 1-forms and vectors, we can imagine focusing on a very small neighbourhood around some point located among these loci of constant phase. On this very small scale, the loci of constant phase will look flat rather than circular, but the wave vectors $\mathbf{k}$ will still be orthogonal to them. What we do is interpret this local pattern of (flat) surfaces of constant phase as the 1-form $\mathbf{\tilde{k}}$This 1-form corresponding to the wave vector $\mathbf{k}$ is

$\mathbf{\tilde{k}} = \langle \mathbf{k}, \ \rangle$

and as before we interpret this as a function waiting for a vector input. When it receives a vector input, say $\mathbf{v}$, it will output a number computed as the scalar product of $\mathbf{k}$ and $\mathbf{v}$. Thus we can write

$\mathbf{\tilde{k}}(\mathbf{v}) = \langle \mathbf{k}, \mathbf{v} \rangle = \mathbf{k} \cdot \mathbf{v}$

As indicated in the diagram, the vector $\mathbf{v}$ which we supply to $\mathbf{\tilde{k}}$ will be at an angle to the wave vector $\mathbf{k}$. If the vector $\mathbf{v}$ is parallel to the loci of constant phase then $\mathbf{\tilde{k}}(\mathbf{v}) = 0$ because $\mathbf{k}$ and $\mathbf{v}$ will be orthogonal. In the language of 1-forms, this would be interpreted by saying that the vector $\mathbf{v}$ will not pierce the 1-form $\mathbf{\tilde{k}}$ because it will not cross any of the loci of constant phase. Conversely, if the vector $\mathbf{v}$ is parallel to the wave vector $\mathbf{k}$ (orthogonal to the loci of constant phase), we would say that $\mathbf{v}$ will pierce the 1-form $\mathbf{\tilde{k}}$ as much as possible, because it will cross as many loci of constant phase as it possibly can. Between these extremes we will get intermediate values of the 1-form $\mathbf{\tilde{k}}(\mathbf{v})$. The key idea, then, is that the set of loci of constant phase in the neighbourhood of a point is the diagrammatic representation of the 1-form $\mathbf{\tilde{k}}$. When we feed a vector $\mathbf{v}$ into this 1-form we get a measure $\mathbf{\tilde{k}}(\mathbf{v})$ of how many loci of constant phase the vector pierces. This is the language being used by MTW in the prelude to Exercise 2.1 above.

To actually solve Exercise 2.1, begin by recalling from quantum mechanics that a photon’s momentum $\mathbf{p}$ is such that $|\mathbf{p}| = E/c$ where $E = hf$ is the photonic energy and $f$ is the frequency of the oscillator. Since $\lambda f = c$ where $\lambda$ is the photon’s wavelength, we have $E = hc/\lambda$ so the magnitude of the photon’s momentum is

$|\mathbf{p}| = \frac{E}{c} = \frac{h}{\lambda} = \hbar \frac{2\pi}{\lambda} = \hbar |\mathbf{k}|$

and in fact

$\mathbf{p} = \hbar \mathbf{k}$

Note that therefore

$\lambda = \frac{h}{|\mathbf{p}|}$

Famously, de Broglie’s idea in his 1924 PhD thesis was that this wavelength formula applies not just to photons but also to massive particles such as electrons, for which the momentum $\mathbf{p}$ would be calculated as

$\mathbf{p} = m \mathbf{u}$

where $m$ is the mass of the particle and $\mathbf{u}$ is its four-velocity in Minkowski spacetime. Note that this four-velocity is such that $\mathbf{u}\cdot\mathbf{u} = -1$ (easily demonstrated using the $- +++$ metric of Minkowski spacetime).

Thus we have

$\mathbf{p} = m \mathbf{u} = \hbar \mathbf{k}$

so

$\mathbf{u} = \frac{\hbar}{m} \mathbf{k}$

In the prelude to Exercise 2.1, MTW say

relabel the surfaces of $\mathbf{\tilde{k}}$ by $\hbar \times phase$, thereby obtaining the momentum 1-form $\mathbf{\tilde{p}}$. Pierce this 1-form with any vector $\mathbf{v}$, and find the result that $\mathbf{p} \cdot \mathbf{v} = \mathbf{\tilde{p}}(\mathbf{v})$.

Following the authors’ instructions, we relabel the surfaces of $\mathbf{\tilde{k}}$ (i.e., the loci of constant phase) by multiplying by $\hbar$ to get the 1-form

$\mathbf{\tilde{p}} = \hbar \mathbf{\tilde{k}} = \hbar \langle \mathbf{k}, \ \rangle$

As usual, this 1-form is a linear function waiting for a vector input. Supplying the input $\mathbf{v}$ we then get

$\mathbf{\tilde{p}}(\mathbf{v}) = \hbar \langle \mathbf{k}, \mathbf{v} \rangle = \hbar \mathbf{k} \cdot \mathbf{v}$

But this is exactly what we get when we work out $\mathbf{p} \cdot \mathbf{v}$ since

$\mathbf{p} \cdot \mathbf{v} = m \mathbf{u} \cdot \mathbf{v} = m \frac{\hbar}{m} \mathbf{k} \cdot \mathbf{v} = \hbar \mathbf{k} \cdot \mathbf{v}$

Thus, we have solved Exercise 2.1 by showing that $\mathbf{p} \cdot \mathbf{v} = \mathbf{\tilde{p}}(\mathbf{v})$ is in accord with the quantum mechanical properties of de Broglie waves, as claimed by MTW.