Skip to main content
Physics LibreTexts

2.8: Displacement Operators

Consider a system with one degree of freedom corresponding to the Cartesian coordinate \(x\) . Suppose that we displace this system some distance along the \(x\) -axis. We could imagine that the system is on wheels, and we just give it a little push. The final state of the system is completely determined by its initial state, together with the direction and magnitude of the displacement. Note that the type of displacement we are considering is one in which everything to do with the system is displaced. So, if the system is subject to an external potential then the potential must be displaced.

The situation is not so clear with state kets. The final state of the system only determines the direction of the displaced state ket. Even if we adopt the convention that all state kets have unit norms, the final ket is still not completely determined, because it can be multiplied by a constant phase-factor. However, we know that the superposition relations between states remain invariant under the displacement. This follows because the superposition relations have a physical significance that is unaffected by a displacement of the system. Thus, if


$\displaystyle \vert R\rangle = \vert A\rangle + \vert B\rangle$ (194)


in the undisplaced system, and the displacement causes ket $ \vert R\rangle$ to transform to ket $ \vert Rd\rangle$ , etc., then in the displaced system we have


$\displaystyle \vert Rd\rangle = \vert Ad\rangle + \vert Bd\rangle.$ (195)



Incidentally, this determines the displaced kets to within a single arbitrary phase-factor to be multiplied into all of them. The displaced kets cannot be multiplied by individual phase-factors, because this would wreck the superposition relations.

Since Equation (195) holds in the displaced system whenever Equation (194) holds in the undisplaced system, it follows that the displaced ket $ \vert Rd\rangle$ must be the result of some linear operator acting on the undisplaced ket $ \vert R\rangle$ . In other words,


$\displaystyle \vert R d\rangle = D \,\vert R\rangle,$ (196)



where $ D$ an operator that depends only on the nature of the displacement. The arbitrary phase-factor by which all displaced kets may be multiplied results in $ D$ being undetermined to an arbitrary multiplicative constant of modulus unity.

We now adopt the ansatz that any combination of bras, kets, and dynamical variables that possesses a physical significance is invariant under a displacement of the system. The normalization condition


$\displaystyle \langle A\vert A\rangle = 1$ (197)



for a state ket $ \vert A\rangle$ certainly has a physical significance. Thus, we must have


$\displaystyle \langle Ad\vert Ad\rangle = 1.$ (198)



Now, $ \vert Ad\rangle = D\,\vert A\rangle$ and $ \langle Ad\vert = \langle A\vert\,D^{\dag }$ , so


$\displaystyle \langle A\vert\, D^{\dag } \,D\,\vert A\rangle = 1.$ (199)



Because this must hold for any state ket $ \vert A\rangle$ , it follows that


$\displaystyle D^{\dag } \,D = 1.$ (200)



Hence, the displacement operator is unitary. Note that the above relation implies that


$\displaystyle \vert A\rangle = D^{\dag }\, \vert A d\rangle.$ (201)



The equation


$\displaystyle v \,\vert A\rangle = \vert B\rangle,$ (202)



where the operator $ v$ represents a dynamical variable, has some physical significance. Thus, we require that


$\displaystyle v_d\, \vert Ad\rangle = \vert Bd\rangle,$ (203)



where $ v_d$ is the displaced operator. It follows that


$\displaystyle v_d\, \vert Ad\rangle = D\, \vert B\rangle = D \,v \,\vert A\rangle = D\, v\, D^{\dag }\, \vert Ad\rangle.$ (204)



Since this is true for any ket $ \vert Ad\rangle$ , we have


$\displaystyle v_d = D\, v\, D^{\dag }.$ (205)



Note that the arbitrary numerical factor in $ D$ does not affect either of the results (200) and (205).

Suppose, now, that the system is displaced an infinitesimal distance $ \delta x$ along the $ x$ -axis. We expect that the displaced ket $ \vert Ad\rangle$ should approach the undisplaced ket $ \vert A\rangle$ in the limit as $ \delta x\rightarrow 0$ . Thus, we expect the limit


$\displaystyle \lim_{\delta x\rightarrow 0 } \frac{\vert A d\rangle - \vert A\ra...
...}{\delta x} = \lim_{\delta x\rightarrow 0 }\frac{D-1}{\delta x}\,\vert A\rangle$ (206)



to exist. Let


$\displaystyle d_x = \lim_{\delta x\rightarrow 0 }\frac{D-1}{\delta x},$ (207)



where $ d_x$ is denoted the displacement operator along the $ x$ -axis. The fact that $ D$ can be replaced by $ D \,\exp(\,{\rm i}\,\gamma)$ , where $ \gamma$ is a real phase-angle, implies that $ d_x$ can be replaced by


$\displaystyle \lim_{\delta x\rightarrow 0 }\frac{D\,\exp({\rm i}\,\gamma)-1}{\d...
...elta x\rightarrow 0 }\frac{D-1+{\rm i}\, \gamma}{\delta x}= d_x + {\rm i}\,a_x,$ (208)



where $ a_x$ is the limit of $ \gamma/\delta x$ . We have assumed, as seems reasonable, that $ \gamma$ tends to zero as $ \delta x\rightarrow 0$ . It is clear that the displacement operator is undetermined to an arbitrary imaginary additive constant.

For small $ \delta x$ , we have


$\displaystyle D = 1 + \delta x\,d_x.$ (209)



It follows from Equation (200) that


$\displaystyle (1+ \delta x\,d_x^{\,\dag }) \,(1+ \delta x\,d_x) = 1.$ (210)



Neglecting order $ (\delta x)^{\,2}$ , we obtain


$\displaystyle d_x^{~\dag } + d_x = 0.$ (211)



Thus, the displacement operator is anti-Hermitian. Substituting into Equation (205), and again neglecting order $ (\delta x)^2$ , we find that


$\displaystyle v_d = (1+ \delta x\,d_x)\, v\, (1- \delta x\,d_x) = v + \delta x\,( d_x\, v - v\, d_x),$ (212)



which implies


$\displaystyle \lim_{\delta x\rightarrow 0} \frac{v_d -v}{\delta x} = d_x \,v -v\, d_x.$ (213)



Let us consider a specific example. Suppose that a state has a wavefunction $ \psi(x')$ . If the system is displaced a distance $ \delta x$ along the $ x$ -axis then the new wavefunction is $ \psi(x'-\delta x)$ (i.e., the same shape shifted in the $ x$ -direction by a distance $ \delta x$ ). Actually, the new wavefunction can be multiplied by an arbitrary number of modulus unity. It can be seen that the new wavefunction is obtained from the old wavefunction according to the prescription $ x'\rightarrow x'- \delta x$ . Thus,


$\displaystyle x_d = x -\delta x.$ (214)



A comparison with Equation (213), using $ x=v$ , yields


$\displaystyle d_x \,x - x\,d_x = -1.$ (215)



It follows that $ {\rm i}\,\hbar\, d_x$ obeys the same commutation relation with $ x$ that $ p_x$ , the momentum conjugate to $ x$ , does [see Equation (116)]. The most general conclusion we can draw from this observation is that


$\displaystyle p_x = {\rm i}\,\hbar\, d_x + f(x),$ (216)



where $ f$ is Hermitian (since $ p_x$ is Hermitian). However, the fact that $ d_x$ is undetermined to an arbitrary additive imaginary constant (which could be a function of $ x$ ) enables us to transform the function $ f$ out of the above equation, leaving


$\displaystyle p_x = {\rm i}\,\hbar\, d_x.$ (217)



Thus, the displacement operator in the $ x$ -direction is proportional to the momentum conjugate to $ x$ . We say that $ p_x$ is the generator of translations along the $ x$ -axis.

A finite translation along the $ x$ -axis can be constructed from a series of very many infinitesimal translations. Thus, the operator $ D({\mit\Delta} x)$ which translates the system a distance $ {\mit\Delta} x$ along the $ x$ -axis is written


$\displaystyle D({\mit\Delta} x) = \lim_{N\rightarrow \infty} \left(1-{\rm i}\, \frac{{\mit\Delta} x}{N} \frac{p_x}{\hbar}\right)^N,$ (218)



where use has been made of Equations (209) and (217). It follows that


$\displaystyle D({\mit\Delta} x) = \exp\left({-\rm i} \,p_x\,{\mit\Delta} x /\hbar\right).$ (219)



The unitary nature of the operator is now clearly apparent.

We can also construct displacement operators which translate the system along the $ y$ - and $ z$ -axes. Note that a displacement a distance $ {\mit\Delta} x$ along the $ x$ -axis commutes with a displacement a distance $ {\mit\Delta} y$ along the $ y$ -axis. In other words, if the system is moved $ {\mit\Delta} x$ along the $ x$ -axis, and then $ {\mit\Delta} y$ along the $ y$ -axis, then it ends up in the same state as if it were moved $ {\mit\Delta} y$ along the $ y$ -axis, and then $ {\mit\Delta} x$ along the $ x$ -axis. The fact that translations in independent directions commute is clearly associated with the fact that the conjugate momentum operators associated with these directions also commute [see Equations (115) and (219)].