Schrödinger's representation - I

Consider the simple system described in the previous section. A general state ket can be written $\psi(x)\rangle$, where $\psi(x)$ is a general function of the position operator $x$, and $\psi(x')$ is the associated wave-function. Consider the ket whose wave-function is $d\psi(x')/dx'$. This ket is denoted $d\psi/dx\rangle$. The new ket is clearly a linear function of the original ket, so we can think of it as the result of some linear operator acting on $\psi\rangle$. Let us denote this operator $d/dx$. It follows that

$$\frac{d}{dx} \psi\rangle = \frac{d\psi}{dx}\rangle.$$ (127)

Any linear operator which acts on ket vectors can also act on bra vectors. Consider $d/dx$ acting on a general bra $\langle \phi(x)$. According to Eq. (34), the bra $\langle \phi d/dx$ satisfies

$$\left( \langle \phi \frac{d}{dx} \right) \psi\rangle = \langle \phi\left( \frac{d}{dx}\psi\rangle\right).$$ (128)

Making use of Eqs. (117) and (119), we can write

$$\int_{-\infty}^{+\infty} \langle \phi \frac{d}{dx} \vert x'\... ...nt_{-\infty}^{+\infty} \phi(x') dx' \frac{d\psi(x')}{dx'}.$$ (129)

The right-hand side can be transformed via integration by parts to give

$$\int_{-\infty}^{+\infty} \langle \phi \frac{d}{dx} \vert x'\... ...t_{-\infty}^{+\infty} \frac{d \phi(x')}{dx'} dx' \psi(x'),$$ (130)

assuming that the contributions from the limits of integration vanish. It follows that

$$\langle \phi \frac{d}{dx} \vert x'\rangle = - \frac{d \phi(x')}{dx'},$$ (131)

which implies

$$\langle \phi \frac{d}{dx} = - \langle \frac{d\phi}{dx}.$$ (132)

The neglect of contributions from the limits of integration in Eq. (130) is reasonable because physical wave-functions are square-integrable [see Eq. (121)]. Note that

$$\frac{d}{dx} \psi\rangle = \frac{d\psi}{dx}\rangle \stackrel... ...angle \frac{d\psi^\ast}{dx} = -\langle \psi^\ast \frac{d}{dx},$$ (133)

where use has been made of Eq. (132). It follows, by comparison with Eqs. (35) and (126), that

$$\left(\frac{d}{dx}\right)^{\dag }= - \frac{d}{dx}.$$ (134)

Thus, $d/dx$ is an anti-Hermitian operator.

Let us evaluate the commutation relation between the operators $x$ and $d/dx$. We have

$$\frac{d}{dx} x \psi\rangle = \frac{d (x \psi)}{dx}\rangle = x \frac{d}{dx}\psi\rangle + \psi\rangle.$$ (135)

Since this holds for any ket $\psi\rangle$, it follows that

$$\frac{d}{dx} x - x \frac{d}{dx} = 1.$$ (136)

Let $p$ be the momentum conjugate to $x$ (for the simple system under consideration $p$ is a straight-forward linear momentum). According to Eq. (115), $x$ and $p$ satisfy the commutation relation

$$x p - p x = {\rm i} \hbar.$$ (137)

It can be seen, by comparison with Eq. (136), that the Hermitian operator $-{\rm i} \hbar d/dx$ satisfies the same commutation relation with $x$ that $p$ does. The most general conclusion which may be drawn from a comparison of Eqs. (136) and (137) is that

$$p = -{\rm i} \hbar \frac{d}{dx} + f(x),$$ (138)

since (as is easily demonstrated) a general function $f(x)$ of the position operator automatically commutes with $x$.

We have chosen to normalize the eigenkets and eigenbras of the position operator so that they satisfy the normalization condition (116). However, this choice of normalization does not uniquely determine the eigenkets and eigenbras. Suppose that we transform to a new set of eigenbras which are related to the old set via

$$\langle x'\vert _{\rm new} = {\rm e}^{ {\rm i} \gamma'}\langle x'\vert _{\rm old},$$ (139)

where $\gamma'\equiv \gamma(x')$ is a real function of $x'$. This transformation amounts to a rearrangement of the relative phases of the eigenbras. The new normalization condition is

$$\langle x'\vert x''\rangle_{\rm new} = \langle x'\vert {\rm e}^{ {\rm i} \gamma'} {\rm e}^{-{\rm i} \... ... = {\rm e}^{ {\rm i} (\gamma'-\gamma'')} \langle x'\vert x''\rangle_{\rm old}$$

$$= { \rm e}^{ {\rm i} (\gamma'-\gamma'')}\delta(x'-x'')= \delta(x'-x'').$$ (140)

Thus, the new eigenbras satisfy the same normalization condition as the old eigenbras.

By definition, the standard ket $\rangle$ satisfies $\langle x'\vert\rangle = 1$. It follows from Eq. (139) that the new standard ket is related to the old standard ket via

$$\rangle_{\rm new} = {\rm e}^{-{\rm i} \gamma} \rangle_{\rm old},$$ (141)

where $\gamma\equiv \gamma(x)$ is a real function of the position operator $x$. The dual of the above equation yields the transformation rule for the standard bra,

$$\langle_{\rm new} = \langle_{\rm old} {\rm e}^{ {\rm i} \gamma}.$$ (142)

The transformation rule for a general operator $A$ follows from Eqs. (141) and (142), plus the requirement that the triple product $\langle A\rangle$ remain invariant (this must be the case, otherwise the probability of a measurement yielding a certain result would depend on the choice of eigenbras). Thus,

$$A_{\rm new} = {\rm e}^{-{\rm i} \gamma} A_{\rm old} {\rm e}^{ {\rm i} \gamma}.$$ (143)

Of course, if $A$ commutes with $x$ then $A$ is invariant under the transformation. In fact, $d/dx$ is the only operator (we know of) which does not commute with $x$, so Eq. (143) yields

$$\left(\frac{d}{dx}\right)_{\rm new} = {\rm e}^{-{\rm i} \ga... ...\rm i} \gamma} = \frac{d}{dx} + {\rm i} \frac{d\gamma}{dx},$$ (144)

where the subscript ``old'' is taken as read. It follows, from Eq. (138), that the momentum operator $p$ can be written

$$p = -{\rm i} \hbar \left(\frac{d}{dx}\right)_{\rm new} - \hbar \frac{d\gamma}{dx} + f(x).$$ (145)

Thus, the special choice

$$\hbar \gamma(x) = \int^x f(x) dx$$ (146)

yields

$$p = -{\rm i} \hbar \left(\frac{d}{dx}\right)_{\rm new}.$$ (147)

Equation (146) fixes $\gamma$ to within an arbitrary additive constant: i.e., the special eigenkets and eigenbras for which Eq. (147) is true are determined to within an arbitrary common phase-factor.

In conclusion, it is possible to find a set of basis eigenkets and eigenbras of the position operator $x$ which satisfy the normalization condition (116), and for which the momentum conjugate to $x$ can be represented as the operator

$$p = -{\rm i} \hbar \frac{d}{dx}.$$ (148)

A general state ket is written $\psi(x)\rangle$, where the standard ket $\rangle$ satisfies $\langle x'\vert\rangle = 1$, and where $\psi(x')= \langle x'\vert \psi(x)\rangle$ is the wave-function. This scheme of things is known as Schrödinger's representation, and is the basis of wave mechanics.