A Weak Derivation of the Momentum Operator

Here is a cute trick to derive the momentum operator in quantum mechanics. Usually, we taken this for a given or a definition, however there are ways to prove the relationship. Here we explore the relationship between position and momentum using Fourier Transforms.

Suppose that we have an arbitrary wave function $latex \displaystyle |\Psi\rangle\in\mathcal{H}$ represented by the Dirac formulation of non-relativistic canonical quantum mechanics. Now, if we are in the position basis, recall that

$latex \displaystyle \langle x|\Psi\rangle\equiv\Psi(x).$

Let the position operator applied to $latex \displaystyle \Psi(x)$ be given by

$latex \displaystyle \hat{x}\Psi(x)=x_0\Psi(x),$

where $latex \displaystyle x$ is the eigenvalue of the position operator $latex \displaystyle \hat{x}.$ Now let us attempt to understand how the conjugate Fourier Basis, or the momentum basis, can be represented by a function $latex \displaystyle |\Phi\rangle\in\mathcal{H}.$ Similar to the position operator, suppose that the momentum operator on $latex \displaystyle \Phi(p)$ is given by

$latex \displaystyle \hat{p}\Phi(p)=p\Phi(p),$

where $latex \displaystyle p_0$ is the eigenvalue of the momentum operator. An additional relation required for this derivation is the completeness relations for the momentum and position basis, which are

$$\begin{align*}&\int_{-\infty}^{\infty} dp  | p \rangle \langle p | = 1, \text{ and}\\ &\int_{-\infty}^{\infty} dx  | x \rangle \langle x | = 1. \end{align*}$$

Finally, we state the connection between the position and momentum representations which is given by

$latex \displaystyle \langle x|p\rangle = e^{ipx/h}.$

With these relations, we are ready to discover the form of the momentum operator. Quick Note: This can easily be generalized to higher dimensions, there is a wonderful discussion in Zettili’s Quantum Mechanics Concepts and Applications $latex \displaystyle -$ which I believe to provide the best discussion for an undergraduate quantum course.

To find the connection between $latex \displaystyle \hat{p}$ and $latex \displaystyle \hat{x}$, we assume that $latex \displaystyle \Psi(x)=\mathcal{F}^{-1}[\Phi(p)],$ that is

$latex \displaystyle \mathcal{F}^{-1}[\Phi(p)] \equiv \int_{-\infty}^{\infty} e^{ipx/h}\Phi(p)dp. $

Before we continue, we note that the $latex \displaystyle h$ factor is simply a dimensionful constant that is required to make the argument of the exponent unitless. As it turns out from experiment, $latex \displaystyle h$ is Planck’s constant. Returning to the above equation, we have that $latex \displaystyle \Phi(p)=\mathcal{F}[\mathcal{F}^{-1}[\Phi(p)]]=\mathcal{F}[\Psi(x)]$ implying that

$latex \displaystyle \Phi(p) \equiv \frac{1}{2\pi} \int_{-\infty}^{\infty}e^{-ipx/h}\Psi(x)dx.$

Then to find the form of the momentum operator in the position basis, we operate $latex \displaystyle  \hat{p}$ on $latex \displaystyle \Psi(x)$,

$$\begin{align*}\hat{p}\Psi(x) &= \hat{p}\langle x|\Psi \rangle = \langle x|\hat{p}|\Psi \rangle\\ &= \int_{-\infty}^{\infty} \langle x|\hat{p}| p \rangle \langle p |\Psi \rangle dp = \int_{-\infty}^{\infty} \langle x|p| p \rangle \langle p |\Psi \rangle dp\\ &=\int_{-\infty}^{\infty} p\langle x| p \rangle \Phi(p) dp,\end{align*}$$

where we have simply recast $latex \displaystyle \Psi(p)$ as $latex \displaystyle \Phi(p).$ Applying the connection between the different representations, we have that

$$\begin{align*} \displaystyle \hat{p}\Psi(x) &= \int_{-\infty}^{\infty} p\langle x| p \rangle \Phi(p) dp = \int_{-\infty}^{\infty} p e^{ipx/h} \Phi(p) dp \\ &= \int_{-\infty}^{\infty} \left(\frac{h}{i}\frac{\partial}{\partial x} \right) e^{ipx/h} \Phi(p) dp \end{align*}$$

Since $latex \displaystyle \Phi(p)$ has no $latex \displaystyle x$-component, we may remove the derivative operation from the integrand to obtain

$$ \begin{align*} \hat{p}\Psi(x) &= \int_{-\infty}^{\infty} \left(\frac{h}{i}\frac{\partial}{\partial x} \right) e^{ipx/h} \Phi(p) dp \\ &= \left(\frac{h}{i}\frac{\partial}{\partial x} \right) \int_{-\infty}^{\infty} e^{ipx/h} \Phi(p) dp \\ &=  \left(\frac{h}{i}\frac{\partial}{\partial x} \right) \Psi(x), \end{align*}$$

noticing that the integral is simply the Fourier Transform of $latex \displaystyle \Psi(x).$ So, equating both sides of the equation, we have derived the position representation of the momentum operator:

$latex \displaystyle \boxed{\boxed{\hat{p}=\left(\frac{h}{i}\frac{\partial}{\partial x} \right)}}$

Fourier Wave Packet illustrating aspects of the Fourier Space. Image found on Wikipedia. Author is Oleg Alexandrov.

2 thoughts on “A Weak Derivation of the Momentum Operator”

    1. So the exponential should be unit-less. In order to do that, we introduce “h” to have units of [momentum]*[position]. This way the units of the denominator will cancel those of the numerator. If you were to reorganize [momentum]*[position] then you will find the units [Joules]*[seconds].

      I hope this helps!

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