Translation operator from passive transformations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11.

We’ve seen that the translation operator {T\left(\varepsilon\right)} in quantum mechanics can be derived by considering the translation to be an active transformation, that is, a transformation where the state vectors, rather than the operators, get transformed according to

\displaystyle T\left(\varepsilon\right)\left|\psi\right\rangle =\left|\psi_{\varepsilon}\right\rangle \ \ \ \ \ (1)

Using this approach, we found that

\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}P \ \ \ \ \ (2)

 

so that the momentum {P} is the generator of the transformation.

We can also derive {T} using a passive transformation, where the state vectors remain the same but the operators are transformed according to

\displaystyle T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right) \displaystyle = \displaystyle X+\varepsilon I\ \ \ \ \ (3)
\displaystyle T^{\dagger}\left(\varepsilon\right)PT\left(\varepsilon\right) \displaystyle = \displaystyle P \ \ \ \ \ (4)

This is equivalent to an active transformation since

\displaystyle \left\langle \psi\left|T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right)\right|\psi\right\rangle \displaystyle = \displaystyle \left\langle T\left(\varepsilon\right)\psi\left|X\right|T\left(\varepsilon\right)\psi\right\rangle \ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle \left\langle \psi_{\varepsilon}\left|X\right|\psi_{\varepsilon}\right\rangle \ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle x+\varepsilon \ \ \ \ \ (7)

As before we start by taking

\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}G \ \ \ \ \ (8)

where {G} is some Hermitian operator, so that {G^{\dagger}=G}. Plugging this into 3 we get, keeping only terms up to order {\varepsilon}:

\displaystyle T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right) \displaystyle = \displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)X\left(I-\frac{i\varepsilon}{\hbar}G\right)\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle X+\frac{i\varepsilon}{\hbar}I\left(GX-XG\right)\ \ \ \ \ (10)
\displaystyle \displaystyle = \displaystyle X-\frac{i\varepsilon}{\hbar}\left[X,G\right]\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle X+\varepsilon I \ \ \ \ \ (12)

Therefore

\displaystyle -\frac{i\varepsilon}{\hbar}\left[X,G\right] \displaystyle = \displaystyle \varepsilon I\ \ \ \ \ (13)
\displaystyle \left[X,G\right] \displaystyle = \displaystyle i\hbar I \ \ \ \ \ (14)

Since {\left[X,P\right]=i\hbar} we see that

\displaystyle G=P+f\left(X\right) \ \ \ \ \ (15)

The extra {f\left(X\right)} is there because any function of {X} alone commutes with {X}, so

\displaystyle \left[X,G\right]=\left[X,P\right]+\left[X,f\left(X\right)\right]=i\hbar I+0 \ \ \ \ \ (16)

We can eliminate {f\left(X\right)} by considering 4.

\displaystyle T^{\dagger}\left(\varepsilon\right)PT\left(\varepsilon\right) \displaystyle = \displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)P\left(I-\frac{i\varepsilon}{\hbar}G\right)\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle P+\frac{i\varepsilon}{\hbar}I\left(GP-PG\right)\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle P-\frac{i\varepsilon}{\hbar}\left[P,G\right]\ \ \ \ \ (19)
\displaystyle \displaystyle = \displaystyle P \ \ \ \ \ (20)

Thus we must have {\left[P,G\right]=0}, which means that {G} must be a function of {P} alone. This means that the most general form for {f\left(X\right)} is {f\left(X\right)=\mbox{constant}}, but there’s nothing to be gained by adding some non-zero constant to {G}, so we can take {f\left(X\right)=0}. Thus we end up with the same form 2 that we got from the active transformation.

Translational invariance is the condition that the Hamiltonian is unaltered by a translation. In the passive representation this is stated by the condition

\displaystyle T^{\dagger}\left(\varepsilon\right)HT\left(\varepsilon\right)=H \ \ \ \ \ (21)

 

Since translation is unitary, we can apply a theorem that is valid for any operator {\Omega} which can be expanded in powers of {X} and {P}. For any unitary operator {U}, we have

\displaystyle U^{\dagger}\Omega\left(X,P\right)U=\Omega\left(U^{\dagger}XU,U^{\dagger}PU\right) \ \ \ \ \ (22)

This follows because for a unitary operator {U^{\dagger}U=UU^{\dagger}=I} so we can insert the product {UU^{\dagger}} anywhere we like. In particular, we can insert it between each pair of factors in every term of the power series expansion of {\Omega}, for example

\displaystyle U^{\dagger}X^{2}P^{2}U \displaystyle = \displaystyle U^{\dagger}XXPPU\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle U^{\dagger}XUU^{\dagger}XUU^{\dagger}PUU^{\dagger}PU\ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \left(U^{\dagger}XU\right)^{2}\left(U^{\dagger}PU\right)^{2} \ \ \ \ \ (25)

For 21 this means that

\displaystyle T^{\dagger}\left(\varepsilon\right)H\left(X,P\right)T\left(\varepsilon\right)=H\left(X+\varepsilon I,P\right)=H\left(X,P\right) \ \ \ \ \ (26)

As before, this leads to the condition

\displaystyle \left[P,H\right]=0 \ \ \ \ \ (27)

which means that {P} is conserved, according to Ehrenfest’s theorem.

4 thoughts on “Translation operator from passive transformations

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