The momentum representation

Fourier's theorerm (see Sect. 3.9), applied to wave-functions, yields

$$\psi(x,t) = \frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^{\infty} \bar{\psi}(k,t)\,{\rm e}^{\,{\rm i}\,k\,x}\,dk,$$ (182)

$$\bar{\psi}(k,t) = \frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^\infty \psi(x,t)\,{\rm e}^{-{\rm i}\,k\,x}\,dx,$$ (183)

where $k$ represents wave-number. However, $p = \hbar\,k$. Hence, we can also write

$$\psi(x,t) = \frac{1}{\sqrt{2\,\pi\,\hbar}}\int_{-\infty}^{\infty} \phi(p,t)\,{\rm e}^{\,{\rm i}\,p\,x/\hbar}\,dp,$$ (184)

$$\phi(p,t) = \frac{1}{\sqrt{2\,\pi\,\hbar}}\int_{-\infty}^\infty \psi(x,t)\,{\rm e}^{-{\rm i}\,p\,x/\hbar}\,dx,$$ (185)

where $\sqrt{\hbar}\,\phi(p,t)= \bar{\psi}(k,t)$ is the momentum-space equivalent to the real-space wave-function $\psi(x,t)$.

At this stage, it is convenient to introduce a useful function called the Dirac delta-function. This function, denoted $\delta(x)$, was first devised by Paul Dirac, and has the following rather unusual properties: $\delta(x)$ is zero for $x\neq 0$, and is infinite at $x=0$. However, the singularity at $x=0$ is such that

$$\int_{-\infty}^{\infty}\delta(x)\,dx = 1.$$ (186)

The delta-function is an example of what is known as a generalized function: i.e., its value is not well-defined at all $x$, but its integral is well-defined. Consider the integral

$$\int_{-\infty}^{\infty}f(x)\,\delta(x)\,dx.$$ (187)

Since $\delta(x)$ is only non-zero infinitesimally close to $x=0$, we can safely replace $f(x)$ by $f(0)$ in the above integral (assuming $f(x)$ is well-behaved at $x=0$), to give

$$\int_{-\infty}^{\infty}f(x)\,\delta(x)\,dx = f(0)\,\int_{-\infty}^{\infty}\delta(x)\,dx=f(0),$$ (188)

where use has been made of Eq. (186). A simple generalization of this result yields

$$\int_{-\infty}^\infty f(x)\,\delta(x-x_0)\,dx = f(x_0),$$ (189)

which can also be thought of as an alternative definition of a delta-function.

Suppose that $\psi(x) = \delta(x-x_0)$. It follows from Eqs. (185) and (189) that

$$\phi(p) = \frac{{\rm e}^{-{\rm i}\,p\,x_0/\hbar}}{\sqrt{2\,\pi\,\hbar}}.$$ (190)

Hence, Eq. (184) yields the important result

$$\delta(x-x_0)= \frac{1}{2\pi\,\hbar}\int_{-\infty}^{\infty}{\rm e}^{\,{\rm i}\,p\,(x-x_0)/\hbar}\,dp.$$ (191)

Similarly,

$$\delta(p-p_0)= \frac{1}{2\pi\,\hbar}\int_{-\infty}^{\infty}{\rm e}^{\,{\rm i}\,(p-p_0)\,x/\hbar}\,dx.$$ (192)

It turns out that we can just as well formulate quantum mechanics using momentum-space wave-functions, $\phi(p,t)$, as real-space wave-functions, $\psi(x,t)$. The former scheme is known as the momentum representation of quantum mechanics. In the momentum representation, wave-functions are the Fourier transforms of the equivalent real-space wave-functions, and dynamical variables are represented by different operators. Furthermore, by analogy with Eq. (174), the expectation value of some operator $O(p)$ takes the form

$$\langle O\rangle = \int_{-\infty}^{\infty}\phi^\ast(p,t)\,O(p)\,\phi(p,t)\,dp.$$ (193)

Consider momentum. We can write

$$\langle p\rangle = \int_{-\infty}^{\infty} \psi^\ast(x,t)\left(-{\rm i}\,\hbar\, \frac{\partial}{\partial x}\right)\psi(x,t)\,dx$$

$$= \frac{1}{2\pi\,\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\inft... ...hi^\ast(p',t)\,\phi(p,t)\,p\,{\rm e}^{\,{\rm i}\,(p-p')\,x/\hbar}\,dx\,dp\,dp',$$ (194)

where use has been made of Eq. (184). However, it follows from Eq. (192) that

$$\langle p\rangle = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi^\ast(p',t)\,\phi(p,t)\,p\,\delta(p-p')\,dp\,dp'.$$ (195)

Hence, using Eq. (189), we obtain

$$\langle p\rangle = \int_{-\infty}^{\infty}\phi^\ast(p,t)\,p\... ...p,t)\,dp = \int_{-\infty}^{\infty}p\,\vert\phi\vert^{\,2}\,dp.$$ (196)

Clearly, momentum is represented by the operator $p$ in the momentum representation. The above expression also strongly suggests [by comparison with Eq. (140)] that $\vert\phi(p,t)\vert^{\,2}$ can be interpreted as the probability density of a measurement of momentum yielding the value $p$ at time $t$. It follows that $\phi(p,t)$ must satisfy an analogous normalization condition to Eq. (122): i.e.,

$$\int_{-\infty}^{\infty} \vert\phi(p,t)\vert^{\,2}\,dp = 1.$$ (197)

Consider displacement. We can write

$$\langle x\rangle = \int_{-\infty}^{\infty} \psi^\ast(x,t)\,x\,\psi(x,t)\,dx$$ (198)

$$= \frac{1}{2\pi\,\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\inft... ...{\partial}{\partial p}\right){\rm e}^{\,{\rm i}\,(p-p')\,x/\hbar}\,dx\,dp\,dp'.$$

Integration by parts yields

$$\langle x\rangle= \frac{1}{2\pi\,\hbar}\int_{-\infty}^{\inft... ...bar\,\frac{\partial}{\partial p}\right)\phi(p,t)\,dx\,dp\,dp'.$$ (199)

Hence, making use of Eqs. (192) and (189), we obtain

$$\langle x\rangle= \frac{1}{2\pi\,\hbar}\int_{-\infty}^{\inft... ...{\rm i}\,\hbar\,\frac{\partial}{\partial p}\right)\phi(p)\,dp.$$ (200)

Clearly, displacement is represented by the operator

$$x\equiv{\rm i}\,\hbar\,\frac{\partial}{\partial p}$$ (201)

in the momentum representation.

Finally, let us consider the normalization of the momentum-space wave-function $\phi(p,t)$. We have

$$\int_{-\infty}^{\infty} \psi^\ast(x,t)\,\psi(x,t)\,dx = \fra... ...,\phi(p,t)\,{\rm e}^{\,{\rm i}\,(p-p')\,x/\hbar}\,dx\,dp\,dp'.$$ (202)

Thus, it follows from Eqs. (192) and (189) that

$$\int_{-\infty}^{\infty} \vert\psi(x,t)\vert^{\,2}\,dx =\int_{-\infty}^{\infty}\vert\phi(p,t)\vert^{\,2}\,dp.$$ (203)

Hence, if $\psi(x,t)$ is properly normalized [see Eq. (122)] then $\phi(p,t)$, as defined in Eq. (185), is also properly normalized [see Eq. (197)].

The existence of the momentum representation illustrates an important point: i.e., that there are many different, but entirely equivalent, ways of mathematically formulating quantum mechanics. For instance, it is also possible to represent wave-functions as row and column vectors, and dynamical variables as matrices which act upon these vectors.