Translation operators

Consider a translation operator $\hat{T}_\tau$ defined by:

$$\hat{T}_\tau \vert{\bf r}\rangle = \vert{\bf r}+\tau \rangle,$$ (12.5)

which also implies

$$\langle {\bf r} \vert \hat{T}_\tau^\dag = \langle {\bf r}+\tau\vert.$$ (12.6)

Let us consider the scalar product of two generic wavefunctions $\langle \phi \vert \psi \rangle$, and let us insert the product of operators $\hat{T}_\tau^\dag\hat{T}_\tau$ into it, $\langle \phi \vert \hat{T}_\tau^\dag\hat{T}_\tau \vert \psi \rangle$. Inserting resolutions of the identity, and using  and  we obtain:

$$\langle \phi \vert \hat{T}_\tau^\dag\hat{T}_\tau \vert \psi \rang... ...tau^\dag\hat{T}_\tau \vert {\bf r} \rangle \langle {\bf r} \vert \psi \rangle =$$

$$= \int d^3 {\bf r} d^3 {\bf r^\prime} \phi^*({\bf r^\prime})\langle {\bf r^\prime}+\tau\vert {\bf r}+\tau \rangle \psi({\bf r}) =$$

$$= \int d^3 {\bf r} d^3 {\bf r^\prime} \phi^*({\bf r^\prime})\delta({\bf r}-{\bf r^\prime)} \psi({\bf r}) =$$

$$= \int d^3 {\bf r} \phi^*({\bf r}) \psi({\bf r}) = \langle \phi \vert \psi \rangle.$$ (12.7)

Since this is true for any $\vert\phi\rangle$ and $\vert\psi\rangle$ we must have $\hat{T}_\tau^\dag\hat{T}_\tau = \mathds{1}$, and therefore $\hat{T}_\tau$ is a unitary operator. Since $\hat{T}_\tau^{-1} = \hat{T}_{-\tau}$, the effect of the translation operator on a wavefunction $\vert\psi\rangle$ is:

$$\langle {\bf r} \vert\hat{T}_\tau\vert\psi \rangle = \hat{T}_\tau \psi({\bf r}) = \psi({\bf r}-\tau),$$ (12.8)

where the first equality defines the notation for the application of $\hat{T}_\tau$ on $\psi({\bf r})$, and the second equality results from , $\langle {\bf r} \vert \hat{T}_\tau = \langle {\bf r}-\tau\vert$.

Now consider a generic hamiltonian $\hat{H}$ with no special symmetries and consider the commutator $[\hat{H}, \hat{T}_{\bf\tau}]$. For a generic eigenstate of the hamiltonian $\vert\psi_j \rangle$ we have:

$$\langle {\bf r} \vert\hat{T}_{\bf\tau} \hat{H} \vert\psi_j \rangle = \epsilon_j \psi_j({\bf r}-{\bf\tau}).$$ (12.9)

Let us now compute the second part of the commutator:

$$\langle {\bf r} \vert \hat{H} \hat{T}_{\bf\tau} \vert\psi_j \rangle.$$ (12.10)

To make this more explicit, let us insert resolutions of the identity before $\hat{H}$ and between $\hat{H}$ and $\hat{T}_{\bf\tau}$. We have:

$$\langle {\bf r} \vert \sum_i \vert\psi_i \rangle \langle \psi_i \... ...e} \rangle \langle {\bf r^\prime} \vert \hat{T}_{\bf\tau} \vert\psi_j \rangle =$$

$$\sum_i \epsilon_i \langle {\bf r} \vert\psi_i \rangle \int d^3{\b... ...e} \rangle \langle {\bf r^\prime} \vert \hat{T}_{\bf\tau} \vert\psi_j \rangle =$$

$$\sum_i \epsilon_i \psi_i({\bf r}) \int d^3{\bf r^\prime} \psi_i^* ({\bf r^\prime}) \psi_j( {\bf r^\prime} - \tau).$$ (12.11)

Since there is no special relation between the eigenstates of the hamiltonian $\psi_i$ and $\psi_j$ evaluated at positions differing by $\tau$, we cannot simplify  any further, and so in general  and  are not equal and therefore $\hat{H}$ and $\hat{T}_\tau$ do not commute.