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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

$\displaystyle \vert R\rangle = \vert A\rangle + \vert B\rangle$ (2.103)

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

$\displaystyle \vert Rd\rangle = \vert Ad\rangle + \vert Bd\rangle.$ (2.104)

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.

Given that Equation (2.104) holds in the displaced system whenever Equation (2.103) 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,

$\displaystyle \vert R d\rangle = D \,\vert R\rangle,$ (2.105)

where $ D$ is 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

$\displaystyle \langle A\vert A\rangle = 1$ (2.106)

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

$\displaystyle \langle Ad\vert Ad\rangle = 1.$ (2.107)

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

$\displaystyle \langle A\vert\, D^{\dag } \,D\,\vert A\rangle = 1.$ (2.108)

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

$\displaystyle D^{\dag } \,D = 1.$ (2.109)

Hence, the operator $ D$ is unitary. Note that the previous relation implies that

$\displaystyle \vert A\rangle = D^{\dag }\, \vert A d\rangle.$ (2.110)

The equation

$\displaystyle v \,\vert A\rangle = \vert B\rangle,$ (2.111)

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

$\displaystyle v_d\, \vert Ad\rangle = \vert Bd\rangle,$ (2.112)

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

$\displaystyle v_d\, \vert Ad\rangle = D\, \vert B\rangle = D \,v \,\vert A\rangle = D\, v\, D^{\dag }\, \vert Ad\rangle.$ (2.113)

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

$\displaystyle v_d = D\, v\, D^{\dag }.$ (2.114)

Note that the arbitrary multiplicative factor in $ D$ does not affect either of the results (2.109) or (2.114).

Suppose, now, that the system is displaced an infinitesimal distance $ \delta x$ along the $ x$ -axis. Let $ D_x(\delta x)$ be the operator that accomplishes this displacement. 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

$\displaystyle \lim_{\delta x\rightarrow 0 } \frac{\vert A d\rangle - \vert A\ra...
...= \lim_{\delta x\rightarrow 0 }\frac{D_x(\delta x)-1}{\delta x}\,\vert A\rangle$ (2.115)

to exist. Let

$\displaystyle d_x = \lim_{\delta x\rightarrow 0 }\frac{D_x(\delta x)-1}{\delta x},$ (2.116)

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

$\displaystyle \lim_{\delta x\rightarrow 0 }\frac{D_x(\delta x)\,\exp(\,{\rm i}\...
...arrow 0 }\frac{D_x(\delta x)-1+{\rm i}\, \gamma}{\delta x}= d_x + {\rm i}\,a_x,$ (2.117)

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

$\displaystyle D_x(\delta x) = 1 + \delta x\,d_x.$ (2.118)

It follows from Equation (2.109) that

$\displaystyle (1+ \delta x\,d_x^{\,\dag }) \,(1+ \delta x\,d_x) = 1.$ (2.119)

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

$\displaystyle d_x^{~\dag } + d_x = 0.$ (2.120)

Thus, the displacement operator is anti-Hermitian. Substituting into Equation (2.114), and again neglecting order $ (\delta x)^2$ , we find that

$\displaystyle v_d = (1+ \delta x\,d_x)\, v\, (1- \delta x\,d_x) = v + \delta x\,( d_x\, v - v\, d_x),$ (2.121)

which implies that

$\displaystyle \lim_{\delta x\rightarrow 0} \frac{v_d -v}{\delta x} = d_x \,v -v\, d_x.$ (2.122)

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 function 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,

$\displaystyle x_d = x -\delta x.$ (2.123)

A comparison with Equation (2.122), using $ x=v$ , yields

$\displaystyle d_x \,x - x\,d_x = -1.$ (2.124)

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 (2.25).] The most general conclusion we can draw from this observation is that

$\displaystyle p_x = {\rm i}\,\hbar\, d_x + f(x),$ (2.125)

where $ f(x)$ is Hermitian (because $ 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(x)$ out of the previous equation, leaving

$\displaystyle p_x = {\rm i}\,\hbar\, d_x.$ (2.126)

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 translation along the $ x$ -axis.

A finite displacement along the $ x$ -axis can be constructed from a series of very many infinitesimal displacements. Thus, the operator $ D_x({\mit\Delta} x)$ , which displaces the system a finite distance $ {\mit\Delta} x$ along the $ x$ -axis, is written [1]

$\displaystyle D_x({\mit\Delta} x) = \lim_{N\rightarrow \infty} \left(1-{\rm i}\...
...^N \equiv \sum_{n=0,\infty} \frac{(-{\rm i}\,p_x\,{\mit\Delta} x/\hbar)^n}{n!},$ (2.127)

where use has been made of Equations (2.118) and (2.126). It follows that

$\displaystyle D_x({\mit\Delta} x) = \exp\left(\frac{{-\rm i} \,p_x\,{\mit\Delta} x}{\hbar}\right).$ (2.128)

The unitary nature of the operator is now clearly apparent. (See Exercise 15.)

We can also construct operators that displace the system along the $ y$ - and $ z$ -axes. For instance, the operator that displaces the system a finite distance $ {\mit\Delta}y$ along the $ y$ -axis is

$\displaystyle D_y({\mit\Delta} y) = \exp\left(\frac{{-\rm i} \,p_y\,{\mit\Delta} y}{\hbar}\right).$ (2.129)

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 a physical 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 finite translations in independent directions commute implies that the associated displacement operators also commute. For instance, $ D_x({\mit\Delta x})\,D_y({\mit\Delta y}) = D_y({\mit\Delta y})\,D_x({\mit\Delta x})$ . This property of displacement operators is clearly associated with the fact that the corresponding momentum operators also commute. In this case, $ p_x\,p_y=p_y\,p_x$ . [See Equations (2.24), (2.128), and Exercise 3.]

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Next: Exercises Up: Position and Momentum Previous: Heisenberg Uncertainty Principle
Richard Fitzpatrick 2016-01-22