The translation operator ${\displaystyle {\hat {T}}(\mathbf {x} )}$  moves particles and fields by the amount ${\displaystyle \mathbf {x} }$ . Therefore, if a particle is in an eigenstate ${\displaystyle |\mathbf {r} \rangle }$  of the position operator (i.e., precisely located at the position ${\displaystyle \mathbf {r} }$ ), then after ${\displaystyle {\hat {T}}(\mathbf {x} )=\int \!d\mathbf {r} ~|\mathbf {r+x} \rangle \langle \mathbf {r} |}$  acts on it, the particle is at the position ${\displaystyle \mathbf {r} +\mathbf {x} }$ :

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 ${\displaystyle \psi (\mathbf {r} )}$ , and ${\displaystyle {\hat {T}}(\mathbf {x} )}$  acts on the particle, the new position-space wavefunction is ${\displaystyle \psi '(\mathbf {r} )={\hat {T}}(\mathbf {x} )\psi (\mathbf {r} )}$  defined by

This relation is easier to remember as ${\displaystyle \psi '(\mathbf {r} +\mathbf {x} )=\psi (\mathbf {r} ),}$  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".^[1]

Here is an example showing that these two descriptions are equivalent. The state ${\displaystyle |\mathbf {a} \rangle }$  corresponds to the wavefunction ${\displaystyle \psi (\mathbf {r} )=\delta (\mathbf {r} -\mathbf {a} )}$  (where ${\displaystyle \delta }$  is the Dirac delta function), while the state ${\displaystyle {\hat {T}}(\mathbf {x} )|\mathbf {a} \rangle =|\mathbf {a} +\mathbf {x} \rangle }$  corresponds to the wavefunction ${\displaystyle \psi '(\mathbf {r} )=\delta (\mathbf {r} -(\mathbf {a} +\mathbf {x} )).}$  These indeed satisfy ${\displaystyle \psi '(\mathbf {r} )=\psi (\mathbf {r} -\mathbf {x} ).}$ 

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 counterexample is a charged particle in a magnetic field.^[1] This definition of momentum is especially important because the law of conservation of momentum applies only to canonical momentum, and is not universally valid if momentum is defined instead as mass times velocity (the so-called "kinetic momentum"), for reasons explained below.

The (canonical) momentum operator is defined as the gradient of the translation operators near the origin:

where ${\displaystyle \hbar }$  is the reduced Planck's constant. For example, what is the result when the ${\displaystyle {\hat {p}}_{x}}$  operator acts on a quantum state? To find the answer, translate the state by an infinitesimal amount in the ${\displaystyle x}$ -direction, and calculate the rate that the state is changing, and multiply it by ${\displaystyle i\hbar }$ . For example, if a state does not change at all when it is translated in the ${\displaystyle x}$ -direction, then its ${\displaystyle x}$ -component of momentum is 0.

More explicitly, ${\displaystyle \mathbf {\hat {p}} }$  is a vector operator (i.e. a vector consisting of three operators ${\displaystyle ({\hat {p}}_{x},{\hat {p}}_{y},{\hat {p}}_{z})}$ ), defined by:

The equation above is the most general definition of ${\displaystyle \mathbf {\hat {p}} }$ . In the special case of a single particle with wavefunction ${\displaystyle \psi (\mathbf {r} )}$ , ${\displaystyle \mathbf {\hat {p}} }$  can be written in a more specific and useful form. In one dimension:

We have now defined ${\displaystyle \mathbf {\hat {p}} }$  in terms of translation operators. It is also possible to write a translation operator as a function of ${\displaystyle \mathbf {\hat {p}} }$ . The method consists in expressing a given translation as a huge number ${\displaystyle N}$  of consecutive tiny translations, and then use the fact that infinitesimal translations can be written in terms of ${\displaystyle \mathbf {\hat {p}} }$ :

where ${\displaystyle \exp }$  is the operator exponential and the right-hand side is the Taylor series expansion. For very small ${\displaystyle \mathbf {x} }$ , one can use the approximation:

Hence, the momentum operator is referred to as the generator of translation.^[2]

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:

Successive translations edit

In other words, if particles and fields are moved by the amount ${\displaystyle \mathbf {x} _{2}}$  and then by the amount ${\displaystyle \mathbf {x} _{1}}$ , overall they have been moved by the amount ${\displaystyle \mathbf {x} _{1}+\mathbf {x} _{2}}$ . For a mathematical proof, one can look at what these operators do to a particle in a position eigenstate:

Inverse edit

The translation operators are invertible, and their inverses are:

This follows from the "successive translations" property above, and the fact that ${\displaystyle {\hat {T}}(\mathbf {0} )={\hat {\mathbb {I} }}}$ , i.e. a translation by a distance of 0 is the same as the identity operator which leaves all states unchanged.

Translation operators commute with each other edit

Translation operators are unitary edit

If ${\displaystyle \psi (\mathbf {r} )}$  and ${\displaystyle \phi (\mathbf {r} )}$  are two position-space wavefunctions, then the inner product of ${\displaystyle \psi }$  with ${\displaystyle \phi }$  is:

The fact that translation operators are unitary implies that the momentum operator is Hermitian.^[1]

Translation Operator operating on a bra edit

A translation operator ${\displaystyle {\hat {T}}(\mathbf {x} )=\int d\mathbf {r} ~|\mathbf {r+x} \rangle \langle \mathbf {r} |}$  operating on a bra in the position eigenbasis gives:

Splitting a translation into its components edit

According to the "successive translations" property above, a translation by the vector ${\displaystyle \mathbf {x} =(x,y,z)}$  can be written as the product of translations in the component directions:

Commutator with position operator edit

Suppose ${\displaystyle |\mathbf {r} \rangle }$  is an eigenvector of the position operator ${\displaystyle \mathbf {\hat {r}} }$  with eigenvalue ${\displaystyle \mathbf {r} }$ . We have

Therefore, the commutator between a translation operator and the position operator is:

Commutator with momentum operator edit

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. ${\displaystyle {\hat {p}}_{y}=\lim _{\varepsilon \to 0}{\frac {i\hbar }{\varepsilon }}\left({\hat {T}}((0,\varepsilon ,0))-{\hat {T}}((0,0,0))\right)}$ ), it follows that translation operators all commute with the momentum operator, i.e.

The set ${\displaystyle {\mathfrak {T}}}$  of translation operators ${\displaystyle {\hat {T}}(\mathbf {x} )}$  for all ${\displaystyle \mathbf {x} }$ , 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 ${\displaystyle \mathbf {0} }$  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 ${\displaystyle {\hat {T}}(\mathbf {x} )}$  is the inverse of the reverse translation ${\displaystyle {\hat {T}}(-\mathbf {x} )}$ .

Associativity

This is the claim that ${\displaystyle {\hat {T}}(\mathbf {x} _{1})\left({\hat {T}}(\mathbf {x} _{2}){\hat {T}}(\mathbf {x} _{3})\right)=\left({\hat {T}}(\mathbf {x} _{1}){\hat {T}}(\mathbf {x} _{2})\right){\hat {T}}(\mathbf {x} _{3})}$ . It is true by definition, as is the case for any group based on function composition.

Therefore, the set ${\displaystyle {\mathfrak {T}}}$  of translation operators ${\displaystyle {\hat {T}}(\mathbf {x} )}$  for all ${\displaystyle \mathbf {x} }$  forms a group.^[3] 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.^[4]

The translation group acting on the Hilbert space of position eigenstates is isomorphic to the group of vector additions in the Euclidean space.

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^[5] play the role of the classical variables. For example, if a particle is in a state ${\displaystyle |\psi \rangle }$ , then the expectation value of the position is ${\displaystyle \langle \psi |\mathbf {\hat {r}} |\psi \rangle }$ , where ${\displaystyle \mathbf {\hat {r}} }$  is the position operator.

If a translation operator ${\displaystyle {\hat {T}}(\mathbf {x} )}$  acts on the state ${\displaystyle |\psi \rangle }$ , creating a new state ${\displaystyle |\psi _{2}\rangle }$  then the expectation value of position for ${\displaystyle |\psi _{2}\rangle }$  is equal to the expectation value of position for ${\displaystyle |\psi \rangle }$  plus the vector ${\displaystyle \mathbf {x} }$ . This result is consistent with what you would expect from an operation that shifts the particle by that amount.

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.

In quantum mechanics, the Hamiltonian represents the energy and dynamics of a system. Let ${\displaystyle |\mathbf {r} _{T}\rangle \equiv {\hat {T}}|\mathbf {r} \rangle }$ be a newly translated state (the argument of ${\displaystyle {\hat {T}}(\mathbf {x} )}$  is irrelevant here and is temporarily dropped for brevity). A Hamiltonian is said to be invariant if

Thus, if the Hamiltonian is invariant under translation, the Hamiltonian commutes with the translation operator (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 edit

First we consider the case where all the translation operators are symmetries of the system. As we will see, in this case conservation of momentum occurs.

For example, if ${\displaystyle {\hat {H}}}$  is the Hamiltonian describing all particles and fields in the universe, and ${\displaystyle {\hat {T}}(\mathbf {x} )}$  is the translation operator that shifts all particles and fields in the universe simultaneously by the same amount, then this is always a symmetry: ${\displaystyle {\hat {H}}}$  describes the complete laws of physics in our universe, which are independent of location. As a consequence, conservation of momentum is universally valid.

On the other hand, perhaps ${\displaystyle {\hat {H}}}$  and ${\displaystyle {\hat {T}}(\mathbf {x} )}$  refer to just one particle. Then the translation operators ${\displaystyle {\hat {T}}(\mathbf {x} )}$  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), but it is conserved if the particle is alone in a vacuum.

Since the Hamiltonian commutes with the translation operator when the translation is invariant

Discrete translational symmetry edit

There is another special case where the Hamiltonian may be translationally invariant. This type of translational symmetry is observed whenever the potential is periodic:^[6]

But, whenever ${\displaystyle x_{j}}$  coincides with the period of the potential ${\displaystyle a}$ ,

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 ${\displaystyle V(\mathbf {r} )}$  with the periodicity of the underlying Bravais lattice

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^[7] (in contrast to free particles, to which Bloch electrons reduce when the periodic potential is identically zero.)

For each Bravais lattice vector ${\displaystyle \mathbf {R} }$  we define a translation operator ${\displaystyle {\hat {T}}_{\mathbf {R} }}$  which, when operating on any function ${\displaystyle f(\mathbf {r} )}$  shifts the argument by ${\displaystyle \mathbf {R} }$ :

The eigenvalues ${\displaystyle c(\mathbf {R} )}$  of the translation operators are related because of the condition:

Therefore, one can choose the simultaneous eigenstates ${\displaystyle \psi }$  of the Hamiltonian ${\displaystyle {\hat {H}}}$  and ${\displaystyle {\hat {T}}_{\mathbf {R} }}$  so that for every Bravais lattice vector ${\displaystyle \mathbf {R} }$ ,

This result is known as Bloch's Theorem.

In the passive transformation picture, translational invariance requires,

Example edit

Suppose at ${\displaystyle t=0}$  two observers A and B prepare identical systems at ${\displaystyle x=0}$  and ${\displaystyle x=a}$  (fig. 1), respectively. If ${\displaystyle |\psi (0)\rangle }$  be the state vector of the system prepared by A, then the state vector of the system prepared by B will be given by

• Bloch state
• Group
• Periodic function
• Shift operator
• Symmetries in quantum mechanics
• Time translation symmetry
• Translational symmetry