Translation operator (quantum mechanics)

The translation operator \hat{T}(\mathbf{x}) moves particles and fields by the amount \mathbf x. Therefore, if a particle is in an eigenstate |\mathbf{r}\rangle of the position operator (i.e., precisely located at the position \mathbf{r}), then after \hat{T}(\mathbf{x})= \int\! d\mathbf{r} ~|\mathbf{r+x}\rangle \langle \mathbf{r} | acts on it, the particle is at the position \mathbf{r} + \mathbf{x}: \hat{T}(\mathbf{x})|\mathbf{r}\rangle = |\mathbf{r}+\mathbf{x}\rangle . An alternative (and equivalent) way to describe what the translation operator determines is based on position-space wavefunctions. If a particle has a position-space wavefunction \psi(\mathbf{r}), and \hat T(\mathbf{x}) acts on the particle, the new position-space wavefunction is \psi' (\mathbf{r})= \hat T(\mathbf x) \psi(\mathbf{r}) defined by \psi'(\mathbf{r}) = \psi(\mathbf{r} - \mathbf{x}). This relation is easier to remember as \psi'(\mathbf{r}+\mathbf{x}) = \psi(\mathbf{r}), which can be read as: "The value of the new wavefunction at the new point equals the value of the old wavefunction at the old point". Here is an example showing that these two descriptions are equivalent. The state |\mathbf{a}\rangle corresponds to the wavefunction \psi(\mathbf{r}) = \delta(\mathbf{r} - \mathbf{a}) (where \delta is the Dirac delta function), while the state \hat{T}(\mathbf{x})|\mathbf{a}\rangle = |\mathbf{a}+\mathbf{x}\rangle corresponds to the wavefunction \psi'(\mathbf{r}) = \delta(\mathbf{r} - (\mathbf{a} + \mathbf{x})). These indeed satisfy \psi'(\mathbf{r}) = \psi(\mathbf{r} - \mathbf{x}) . == Momentum as generator of translations ==

In introductory physics, momentum is usually defined as mass times velocity. However, there is a more fundamental way to define momentum, in terms of translation operators. This is more specifically called canonical momentum, and it is usually but not always equal to mass times velocity. One notable exception pertains to a charged particle in a magnetic field in which the canonical momentum includes both the usual momentum and a second term proportional to the magnetic vector potential. A nice way to double-check that these relations are correct is to do a Taylor expansion of the translation operator acting on a position-space wavefunction. Expanding the exponential to all orders, the translation operator generates exactly the full Taylor expansion of a test function: \begin{align} \psi(\mathbf{r}-\mathbf{x}) & =\hat T(\mathbf x)\psi(\mathbf{r})\\ & =\exp\left(-\frac{i\mathbf x\cdot\mathbf{\hat p}}{\hbar}\right)\psi(\mathbf{r})\\ & =\left(\sum_{n=0}^{\infty} \frac{1}{n!}(-\frac{i}{\hbar}\mathbf{x}\cdot\mathbf{\hat{p}})^n\right)\psi(\mathbf{r})\\ & =\left(\sum_{n=0}^{\infty} \frac{1}{n!}(-\mathbf{x}\cdot\mathbf{\nabla})^n\right)\psi(\mathbf{r})\\ & =\psi(\mathbf{r})-\mathbf{x}\cdot\mathbf{\nabla}\psi(\mathbf{r})+\frac{1}{2!}(\mathbf{x}\cdot\mathbf{\nabla})^2\psi(\mathbf{r})-\dots \end{align} So every translation operator generates exactly the expected translation on a test function if the function is analytic in some domain of the complex plane. == Properties ==

Successive translations \hat T(\mathbf{x}_1) \hat T(\mathbf{x}_2) = \hat T(\mathbf{x}_1 + \mathbf{x}_2) In other words, if particles and fields are moved by the amount \mathbf{x}_2 and then by the amount \mathbf{x}_1, overall they have been moved by the amount \mathbf{x}_1 + \mathbf{x}_2. For a mathematical proof, one can look at what these operators do to a particle in a position eigenstate: \hat T(\mathbf{x}_1) \hat T(\mathbf{x}_2) |\mathbf{r}\rangle = \hat T(\mathbf{x}_1) |\mathbf{x}_2 + \mathbf{r}\rangle = |\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{r}\rangle = \hat T(\mathbf{x}_1 + \mathbf{x}_2)|\mathbf{r}\rangle Since the operators \hat T(\mathbf{x}_1) \hat T(\mathbf{x}_2) and \hat T(\mathbf{x}_1 + \mathbf{x}_2) have the same effect on every state in an eigenbasis, it follows that the operators are equal. Identity translation The translation \hat T(\mathbf{0}) = \hat{\mathbb{I}}, i.e. a translation by a distance of 0 is the same as the identity operator which leaves all states unchanged. Inverse The translation operators are invertible, and their inverses are: (\hat T(\mathbf{x}))^{-1} = \hat T(-\mathbf{x}) This follows from the "successive translations" property above, and the identity translation. Translation operators commute with each other \hat T(\mathbf{x})\hat T(\mathbf{y}) = \hat T(\mathbf{y})\hat T(\mathbf{x}) because both sides are equal to \hat T(\mathbf{x}+\mathbf{y}). Translation operators are unitary To show that translation operators are unitary, we first must prove that the momentum operator \widehat{p} is Hermitian. Then, we can prove that the translation operator meets two criteria that are necessary to be a unitary operator. To begin with, the linear momentum operator \widehat{p} : L^2([-\infty,\infty],\mu)\to L^2([-\infty,\infty],\mu) is the rule that assigns to any \psi(r) in the domain the one vector \psi'(r) = \left[\widehat{p}\psi\right](r) = \left[-i\hbar \frac{d}{dx}\psi\right](r) in the codomain is. Since \langle \widehat{p}\psi,\phi\rangle =\langle \psi,\widehat{p}\phi\rangle ,\quad\text{for all}~\psi,\phi\in L^2([-\infty,\infty],\mu) therefore the linear momentum operator \widehat{p} is, in fact, a Hermitian operator. Detailed proofs of this can be found in many textbooks and online (e.g. https://physics.stackexchange.com/a/832341/194354). Having in hand that the momentum operator is Hermitian, we can prove that the translation operator is a unitary operator. First, it must shown that translation operator is a bounded operator. It is sufficient to state that for all \psi\in L^2([a,b],\mu) that \left\|\left[\widehat{T}_{\mathbf{x}}\psi\right]\!(r)\left\|_{L^2([a,b],\mu)} = \right\|\psi(r)\right\|_{L^2([a,b],\mu)}. Second, it must be (and can be) shown that \widehat{T}^\dagger_{\mathbf{x}} \widehat{T} _{\mathbf{x}} = \widehat{T}_{\mathbf{x}} \widehat{T}^\dagger_{\mathbf{x}} = \mathbb{I}. A detailed proof can be found in reference https://math.stackexchange.com/a/4990451/309209. Translation Operator operating on a bra A translation operator \hat{T}(\mathbf{x})= \int d\mathbf{r} ~ |\mathbf{r+x}\rangle \langle \mathbf{r} | operating on a bra in the position eigenbasis gives: \langle \mathbf r| \hat T(\mathbf x) = \langle \mathbf r - \mathbf x| {{math proof|proof= \hat T(\mathbf x)|\mathbf r\rangle = |\mathbf r + \mathbf x\rangle Its adjoint expression is: \langle \mathbf r|(\hat T(\mathbf{x}))^\dagger = \langle \mathbf r + \mathbf x| Using the results above, (\hat T(\mathbf{x}))^\dagger = (\hat T(\mathbf{x}))^{-1} = \hat{T}(-\mathbf x): \langle \mathbf r| \hat T(-\mathbf x) = \langle \mathbf r + \mathbf x| Replacing \mathbf x by -\mathbf x, \langle \mathbf r| \hat T(\mathbf x) = \langle \mathbf r - \mathbf x| }} Splitting a translation into its components According to the "successive translations" property above, a translation by the vector \mathbf{x} = (x,y,z) can be written as the product of translations in the component directions: \hat{T}(\mathbf{x}) = \hat{T}(x\mathbf{\hat{x}})\,\hat{T}(y\mathbf{\hat{y}})\,\hat{T}(z\mathbf{\hat{z}}) where \mathbf{\hat{x}},\mathbf{\hat{y}},\mathbf{\hat{z}} are unit vectors. Commutator with position operator Suppose |\mathbf r\rangle is an eigenvector of the position operator \mathbf{\hat r} with eigenvalue \mathbf r. We have \hat T(\mathbf{x})\mathbf{\hat{r}}|\mathbf r\rangle = \hat T(\mathbf{x})\mathbf{r}|\mathbf r\rangle = \mathbf{r}|\mathbf{x} + \mathbf r\rangle while \mathbf{\hat{r}}\hat T(\mathbf{x})|\mathbf r\rangle = \hat\mathbf{r}|\mathbf{x} + \mathbf r\rangle = (\mathbf{x} + \mathbf{r}) |\mathbf{x} + \mathbf r\rangle Therefore, the commutator between a translation operator and the position operator is: [\mathbf{\hat r},\hat T(\mathbf x)] \equiv \mathbf{\hat r}\hat T(\mathbf x) - \hat T(\mathbf x)\mathbf{\hat r} = \mathbf x \hat T(\mathbf x) This can also be written (using the above properties) as: (\hat T(\mathbf x))^{-1}\mathbf{\hat r}\hat T(\mathbf x) = \mathbf{\hat r} + \mathbf x\hat{\mathbb{I}} where \hat{\mathbb{I}} is the identity operator. Commutator with momentum operator Since translation operators all commute with each other (see above), and since each component of the momentum operator is a sum of two scaled translation operators (e.g. \hat{p}_y = \lim_{\varepsilon \to 0} \frac{i\hbar}{\varepsilon} \left( \hat{T}((0, \varepsilon, 0)) - \hat{T}((0,0,0))\right)), it follows that translation operators all commute with the momentum operator, i.e. \hat T(\mathbf{x}) \hat{\mathbf{p}} = \hat{\mathbf{p}}\hat T(\mathbf{x}) This commutation with the momentum operator holds true generally even if the system is not isolated where energy or momentum may not be conserved. == Translation group ==

The set \mathfrak{T} of translation operators \hat T(\mathbf x) for all \mathbf x, with the operation of multiplication defined as the result of successive translations (i.e. function composition), satisfies all the axioms of a group: ; Closure : When two translations are done consecutively, the result is a single different translation. (See "successive translations" property above.) ; Existence of identity : A translation by the vector \mathbf{0} is the identity operator, i.e. the operator that has no effect on anything. It functions as the identity element of the group. ; Every element has an inverse : As proven above, any translation operator \hat T(\mathbf{x}) is the inverse of the reverse translation \hat T(-\mathbf x). ; Associativity : This is the claim that \hat T(\mathbf{x}_1)\left (\hat T(\mathbf{x}_2) \hat T(\mathbf{x}_3)\right ) = \left (\hat T(\mathbf{x}_1)\hat T(\mathbf{x}_2)\right )\hat T(\mathbf{x}_3). It is true by definition, as is the case for any group based on function composition. Therefore, the set \mathfrak{T} of translation operators \hat T(\mathbf x) for all \mathbf x forms a group. Since there are continuously infinite number of elements, the translation group is a continuous group. Moreover, the translation operators commute among themselves, i.e. the product of two translation (a translation followed by another) does not depend on their order. Therefore, the translation group is an abelian group. The translation group acting on the Hilbert space of position eigenstates is isomorphic to the group of vector additions in the Euclidean space. == Expectation values of position and momentum in the translated state ==

Consider a single particle in one dimension. Unlike classical mechanics, in quantum mechanics a particle neither has a well-defined position nor a well-defined momentum. In the quantum formulation, the expectation values play the role of the classical variables. For example, if a particle is in a state |\psi\rangle, then the expectation value of the position is \langle \psi | \mathbf{\hat r} | \psi \rangle, where \mathbf{\hat r} is the position operator. If a translation operator \hat T(\mathbf{x}) acts on the state |\psi\rangle, creating a new state |\psi_2\rangle then the expectation value of position for |\psi_2\rangle is equal to the expectation value of position for |\psi\rangle plus the vector \mathbf{x}. This result is consistent with what you would expect from an operation that shifts the particle by that amount. {{math proof|title=Proof that a translation operator changes the expectation value of position in the way you would expect|proof=Assume |\psi_2\rangle = \hat T(\mathbf{x})|\psi\rangle as stated above. \begin{align} \langle \psi_2 | \hat{\mathbf{r}} | \psi_2 \rangle &= (\langle \psi | (\hat{T}(\mathbf{x}))^\dagger) \hat{\mathbf{r}} (\hat{T}(\mathbf{x})| \psi \rangle) \\ &= \langle \psi | \hat{\mathbf{r}} | \psi \rangle + \mathbf{x} \end{align} using the normalization condition \langle\psi|\psi\rangle = 1, and the commutator result proven in a previous section. }} On the other hand, when the translation operator acts on a state, the expectation value of the momentum is not changed. This can be proven in a similar way as the above, but using the fact that translation operators commute with the momentum operator. This result is again consistent with expectations: translating a particle does not change its velocity or mass, so its momentum should not change. == Translational invariance ==

In quantum mechanics, the Hamiltonian is the operator corresponding to the total energy of a system. For any |\mathbf{r}\rangle in the domain, let the one vector |\mathbf{r}_T\rangle\equiv \hat{T}_\mathbf{x}|\mathbf{r}\rangle in the codomain be a newly translated state. If \langle \mathbf{r}_T|\hat{H}|\mathbf{r}_T\rangle = \langle \mathbf{r}|\hat H|\mathbf{r}\rangle, then a Hamiltonian is said to be invariant. Since the translation operator is a unitary operator, the antecedent can also be written as \langle \mathbf{r}|\hat{T}_\mathbf{x}^{-1}\hat{H}\hat{T}_\mathbf{x}|\mathbf{r}\rangle = \langle \mathbf{r}|\hat H|\mathbf{r}\rangle. Since this holds for any |\mathbf{r}\rangle in the domain, the implication is that \hat{T}_\mathbf{x}^{-1}\hat{H}\hat{T}_\mathbf{x}=\hat {H} or that \hat{H}\hat{T}_\mathbf{x}-\hat{T}_\mathbf{x}\hat {H}=[\hat{H},\hat{T}_\mathbf{x}]=0. Thus, if Hamiltonian commutes with the translation operator, then the Hamiltonian is invariant under translation. Loosely speaking, if we translate the system, then measure its energy, then translate it back, it amounts to the same thing as just measuring its energy directly. Continuous translational symmetry First we consider the case where all the translation operators are symmetries of the system. Second we consider the case where the translation operator is not a symmetries of the system. As we will see, only in the first case does the conservation of momentum occur. For example, let \hat{H} be the Hamiltonian describing all particles and fields in the universe, and let \hat T(\mathbf x) be the continuous translation operator that shifts all particles and fields in the universe simultaneously by the same amount. If we assert the a priori axiom that this translation is a continuous symmetry of the Hamiltonian (i.e., that \hat{H} is independent of location), then, as a consequence, conservation of momentum is universally valid. On the other hand, perhaps \hat{H} and \hat T(\mathbf x) refer to just one particle. Then the translation operators \hat T(\mathbf x) are exact symmetries only if the particle is alone in a vacuum. Correspondingly, the momentum of a single particle is not usually conserved (it changes when the particle bumps into other objects or is otherwise deflected by the potential energy fields of the other particles), but it is conserved if the particle is alone in a vacuum. Since the Hamiltonian operator commutes with the translation operator when the Hamiltonian is an invariant with respect to translation, therefore \left[\hat H,\hat T(\mathbf{x})\right]=0\,. Further, the Hamiltonian operator also commutes with the infinitesimal translation operator \begin{align} &\left[\hat H, 1-\frac{i\mathbf{x}\cdot \hat\mathbf{p}}{N\hbar}\right]=0\\ & \Rightarrow [\hat H,\hat{\mathbf p}]=0 \\ & \Rightarrow \frac{d}{dt}\langle\hat{\mathbf p}\rangle= \frac{i}{\hbar}[\hat H,\hat{\mathbf p}]=0. \end{align} In summary, whenever the Hamiltonian for a system remains invariant under continuous translation, then the system has conservation of momentum, meaning that the expectation value of the momentum operator remains constant. This is an example of Noether's theorem. Discrete translational symmetry There is another special case where the Hamiltonian may be translationally invariant. This type of translational symmetry is observed whenever the potential is periodic: V(r_j\pm a)=V(r_j) In general, the Hamiltonian is not invariant under any translation represented by \hat T_j(x_j) with x_j arbitrary, where \hat T_j(x_j) has the property: \hat T_j(x_j)|r_j\rangle=|r_j+x_j\rangle and, (\hat{T}_j(x_j))^\dagger \hat r_j\hat T_j(x_j)=\hat r_j+x_j\hat{\mathbb{I}} (where \hat{\mathbb{I}} is the identity operator; see proof above). But, whenever x_j coincides with the period of the potential a, (\hat{T}_j(a))^\dagger V(\hat r_j)\hat T_j(a)=V(\hat r_j+a\hat{\mathbb{I}})=V(\hat r_j) Since the kinetic energy part of the Hamiltonian \hat H is already invariant under any arbitrary translation, being a function of \mathbf{\hat p}, the entire Hamiltonian satisfies, (\hat{T}_j(a))^\dagger \hat H\hat T_j(a)=\hat H Now, the Hamiltonian commutes with translation operator, i.e. they can be simultaneously diagonalised. Therefore, the Hamiltonian is invariant under such translation (which no longer remains continuous). The translation becomes discrete with the period of the potential. == Discrete translation in periodic potential: Bloch's theorem ==

The ions in a perfect crystal are arranged in a regular periodic array. So we are led to the problem of an electron in a potential V(\mathbf{r}) with the periodicity of the underlying Bravais lattice V(\mathbf{r}+\mathbf{R})=V(\mathbf r ) for all Bravais lattice vectors \mathbf R However, perfect periodicity is an idealisation. Real solids are never absolutely pure, and in the neighbourhood of the impurity atoms the solid is not the same as elsewhere in the crystal. Moreover, the ions are not in fact stationary, but continually undergo thermal vibrations about their equilibrium positions. These destroy the perfect translational symmetry of a crystal. To deal with this type of problems the main problem is artificially divided in two parts: (a) the ideal fictitious perfect crystal, in which the potential is genuinely periodic, and (b) the effects on the properties of a hypothetical perfect crystal of all deviations from perfect periodicity, treated as small perturbations. Although, the problem of electrons in a solid is in principle a many-electron problem, in independent electron approximation each electron is subjected to the one-electron Schrödinger equation with a periodic potential and is known as Bloch electron (in contrast to free particles, to which Bloch electrons reduce when the periodic potential is identically zero.) For each Bravais lattice vector \mathbf R we define a translation operator \hat T_{\mathbf R} which, when operating on any function f(\mathbf r) shifts the argument by \mathbf R: \hat T_{\mathbf R}f(\mathbf r)=f(\mathbf r+\mathbf R) Since all translations form an Abelian group, the result of applying two successive translations does not depend on the order in which they are applied, i.e. \hat T_{\mathbf R_1}\hat T_{\mathbf R_2}=\hat T_{\mathbf R_2}\hat T_{\mathbf R_1}=\hat T_{\mathbf {R}_1 + \mathbf{R}_2} In addition, as the Hamiltonian is periodic, we have, \hat T_{\mathbf R}\hat H=\hat H\hat T_{\mathbf R} Hence, the \hat T_{\mathbf R} for all Bravais lattice vectors \mathbf R and the Hamiltonian \hat H form a set of commutating operators. Therefore, the eigenstates of \hat H can be chosen to be simultaneous eigenstates of all the \hat T_{\mathbf R}: \hat H\psi=\mathcal E\psi \hat T_{\mathbf R}\psi=c(\mathbf R)\psi The eigenvalues c(\mathbf R) of the translation operators are related because of the condition: \hat T_{\mathbf R_1}\hat T_{\mathbf R_2}=\hat T_{\mathbf R_2}\hat T_{\mathbf R_1}=\hat T_{\mathbf R_1 + \mathbf R_2} We have, \begin{align} \hat T_{\mathbf R_1}\hat T_{\mathbf R_2}\psi & =c(\mathbf R_1)\hat T_{\mathbf R_2}\psi\\ & =c(\mathbf R_1)c(\mathbf R_2)\psi \end{align} And, \hat T_{\mathbf {R_1 +R_2}}\psi=c(\mathbf R_1+\mathbf R_2)\psi Therefore, it follows that, c(\mathbf R_1+\mathbf R_2)=c(\mathbf R_1)c(\mathbf R_2) Now let the \mathbf{a}_i's be the three primitive vector for the Bravais lattice. By a suitable choice of x_i, we can always write c(\mathbf{a}_i) in the form c(\mathbf{a}_i)=e^{2\pi ix_i} If \mathbf R is a general Bravais lattice vector, given by \mathbf R=n_1\mathbf a_1+n_2\mathbf a_2+n_3\mathbf a_3 it follows then, \begin{align} c(\mathbf R) & =c(n_1\mathbf a_1+n_2\mathbf a_2+n_3\mathbf a_3)\\ & =c(n_1\mathbf a_1)c(n_2\mathbf a_2)c(n_3\mathbf a_3)\\ & =c(\mathbf a_1)^{n_1}c(\mathbf a_2)^{n_2}c(\mathbf a_3)^{n_3} \end{align} Substituting c(\mathbf{a}_i)=e^{2\pi ix_i} one gets, \begin{align} c(\mathbf R) & =e^{2\pi i(n_1x_1+n_2x_2+n_3x_3)}\\ & = e^{i\mathbf k\cdot\mathbf R} \end{align} where \mathbf{k} = x_1 \mathbf{b}_1 + x_2 \mathbf{b}_2 + x_3 \mathbf{b}_3 and the \mathbf b_i's are the reciprocal lattice vectors satisfying the equation \mathbf b_i\cdot\mathbf a_j=2\pi\delta_{ij} Therefore, one can choose the simultaneous eigenstates \psi of the Hamiltonian \hat H and \hat T_{\mathbf R} so that for every Bravais lattice vector \mathbf R, \begin{align} \psi(\mathbf{r+R}) & =\hat T_{\mathbf R}\psi(\mathbf r)\\ & = c(\mathbf R)\psi(\mathbf r)\\ & =e^{i\mathbf k\cdot\mathbf R}\psi(\mathbf r) \end{align} So, {{Equation box 1 This result is known as Bloch's Theorem. == Time evolution and translational invariance ==

In the passive transformation picture, translational invariance requires, [\hat T(\mathbf{x}),\hat H] = 0 It follows that [\hat T(\mathbf{x}),\hat U(t)] = 0 where \hat U(t) is the unitary time evolution operator. When the Hamiltonian is time independent, \hat U(t)= \exp\left(\frac{-i\hat H t}{\hbar}\right) If the Hamiltonian is time dependent, the above commutation relation is satisfied if \hat p or \hat T(\mathbf x) commutes with \hat H (t) for all t. Example Suppose at t=0 two observers A and B prepare identical systems at x=0 and x=a (fig. 1), respectively. If |\psi(0)\rangle be the state vector of the system prepared by A, then the state vector of the system prepared by B will be given by \hat T(\mathbf{a})|\psi(0)\rangle Both the systems look identical to the observers who prepared them. After time t, the state vectors evolve into \hat U (t)|\psi(0)\rangle and \hat U (t) \hat T(\mathbf{a})|\psi(0)\rangle respectively. Using the above-mentioned commutation relation, the later may be written as, \hat T(\mathbf{a})\hat U(t)|\psi(0)\rangle which is just the translated version of the system prepared by A at time t. Therefore, the two systems, which differed only by a translation at t=0, differ only by the same translation at any instant of time. The time evolution of both the systems appear the same to the observers who prepared them. It can be concluded that the translational invariance of Hamiltonian implies that the same experiment repeated at two different places will give the same result (as seen by the local observers). == See also ==