# 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.