The momentum representation

Consider a system with one degree of freedom, describable in terms of a coordinate $x$ and its conjugate momentum $p$, both of which have a continuous range of eigenvalues. We have seen that it is possible to represent the system in terms of the eigenkets of $x$. This is termed Schrödinger's representation. However, it is also possible to represent the system in terms of the eigenkets of $p$.

Consider the eigenkets of $p$ which belong to the eigenvalues $p'$. These are denoted $\vert p'\rangle$. The orthogonality relation for the momentum eigenkets is

$$\langle p'\vert p''\rangle = \delta(p'-p''),$$ (170)

and the corresponding completeness relation is

$$\int_{-\infty}^{+\infty} dp' \vert p'\rangle \langle p'\vert = 1.$$ (171)

A general state ket can be written

$$\phi(p)\rangle$$ (172)

where the standard ket $\rangle$ satisfies

$$\langle p'\vert \rangle = 1.$$ (173)

Note that the standard ket in this representation is quite different to that in Schrödinger's representation. The momentum space wave-function $\phi(p')$ satisfies

$$\phi(p') = \langle p'\vert\phi \rangle.$$ (174)

The probability that a measurement of the momentum yields a result lying in the range $p'$ to $p'+dp'$ is given by

$$P(p', dp') = \vert\phi(p')\vert^2 dp'.$$ (175)

Finally, the normalization condition for a physical momentum space wave-function is

$$\int_{-\infty}^{+\infty} \vert\phi(p')\vert^2 dp' = 1.$$ (176)

The fundamental commutation relations (113)-(115) exhibit a particular symmetry between coordinates and their conjugate momenta. If all the coordinates are transformed into their conjugate momenta, and vice versa, and $\rm i$ is then replaced by $-\rm i$, the commutation relations are unchanged. It follows from this symmetry that we can always choose the eigenkets of $p$ in such a manner that the coordinate $x$ can be represented as (see Sect. 3.4)

$$x = {\rm i} \hbar \frac{d}{dp}.$$ (177)

This is termed the momentum representation.

The above result is easily generalized to a system with more than one degree of freedom. Suppose the system is specified by $N$ coordinates, $q_1\cdots q_N$, and $N$ conjugate momenta, $p_1\cdots p_N$. Then, in the momentum representation, the coordinates can be written as

$$q_i = {\rm i} \hbar \frac{\partial}{\partial p_i}.$$ (178)

We also have

$$q_i \rangle = 0,$$ (179)

and

$$\langle p_1' \cdots p_N'\vert q_i = {\rm i} \hbar \frac{\partial}{\partial p_i'} \langle p_1'\cdots p_N'\vert.$$ (180)

The momentum representation is less useful than Schrödinger's representation for a very simple reason. The energy operator (i.e., the Hamiltonian) of most simple systems takes the form of a sum of quadratic terms in the momenta (i.e., the kinetic energy) plus a complicated function of the coordinates (i.e., the potential energy). In Schrödinger's representation, the eigenvalue problem for the energy translates into a second-order differential equation in the coordinates, with a complicated potential function. In the momentum representation, the problem transforms into a high-order differential equation in the momenta, with a quadratic potential. With the mathematical tools at our disposal, we are far better able to solve the former type of problem than the latter. Hence, Schrödinger's representation is generally more useful than the momentum representation.