Translation operator (quantum mechanics)

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 ${\displaystyle {\hat {T}}(\mathbf {x} )}$ acts on it, the particle is at the position ${\displaystyle \mathbf {r} +\mathbf {x} }$:

$${\displaystyle {\hat {T}}(\mathbf {x} )|\mathbf {r} \rangle =|\mathbf {r} +\mathbf {x} \rangle .}$$

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 ${\displaystyle \psi (\mathbf {r} )}$, 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

${\displaystyle \psi '(\mathbf {r} )=\psi (\mathbf {r} -\mathbf {x} )}$.

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:

${\displaystyle {\hat {p}}_{x}=i\hbar \lim _{a\rightarrow 0}{\frac {{\hat {T}}(a\mathbf {\hat {x}} )-{\hat {\mathbb {I} }}}{a}}}$

where ${\displaystyle {\hat {\mathbb {I} }}}$ is the identity operator and ${\displaystyle \mathbf {\hat {x}} }$ is the unit vector in the ${\displaystyle x}$-direction. (${\displaystyle {\hat {p}}_{y},{\hat {p}}_{z}}$ are defined analogously.)

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:

${\displaystyle {\begin{aligned}({\hat {p}}\psi )(r)&=i\hbar \lim _{a\rightarrow 0}{\frac {({\hat {T}}(a)\psi )(r)-\psi (r)}{a}}\\&=i\hbar \lim _{a\rightarrow 0}{\frac {\psi (r-a)-\psi (r)}{a}}\\&=-i\hbar \left({\frac {\partial \psi }{\partial r}}\right)(r)\end{aligned}}}$

or in three dimensions,

${\displaystyle \mathbf {\hat {p}} =-i\hbar \nabla }$

as an operator acting on position-space wavefunctions. This is the familiar quantum-mechanical expression for ${\displaystyle \mathbf {\hat {p}} }$, but we have derived it here from a more basic starting point.

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 ${\displaystyle \mathbf {\hat {p}} }$:

${\displaystyle {\begin{aligned}{\hat {T}}(\mathbf {x} )&=\lim _{N\to \infty }({\hat {T}}(\mathbf {x} /N))^{N}\\&=\lim _{N\rightarrow \infty }\left[1-{\frac {i\mathbf {x} \cdot \mathbf {\hat {p}} }{N\hbar }}\right]^{N}\end{aligned}}}$

which gives the final expression:

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:

${\displaystyle {\hat {T}}(\mathbf {x} )\approx 1-i\mathbf {x} \cdot \mathbf {\hat {p}} /\hbar }$

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:

${\displaystyle {\begin{aligned}\psi (\mathbf {r} -\mathbf {x} )&={\hat {T}}(\mathbf {x} )\psi (\mathbf {r} )\\&=\exp \left(-{\frac {i\mathbf {x} \cdot \mathbf {\hat {p}} }{\hbar }}\right)\psi (\mathbf {r} )\\&=\left(\sum _{n=0}^{\infty }{\frac {1}{n!}}(-{\frac {i}{\hbar }}\mathbf {x} \cdot \mathbf {\hat {p}} )^{n}\right)\psi (\mathbf {r} )\\&=\left(\sum _{n=0}^{\infty }{\frac {1}{n!}}(-\mathbf {x} \cdot \mathbf {\nabla } )^{n}\right)\psi (\mathbf {r} )\\&=\psi (\mathbf {r} )-\mathbf {x} \cdot \mathbf {\nabla } \psi (\mathbf {r} )+{\frac {1}{2!}}(\mathbf {x} \cdot \mathbf {\nabla } )^{2}\psi (\mathbf {r} )-\dots \end{aligned}}}$

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.

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 you do two translations 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.

Proof that a translation operator changes the expectation value of position in the way you would expect:

Assume ${\displaystyle |\psi _{2}\rangle ={\hat {T}}(\mathbf {x} )|\psi \rangle }$ as stated above. ${\displaystyle {\begin{aligned}\langle \psi _{2}|{\hat {\mathbf {r} }}|\psi _{2}\rangle &=(\langle \psi |({\hat {T}}(\mathbf {x} ))^{\dagger }){\hat {\mathbf {r} }}({\hat {T}}(\mathbf {x} )|\psi \rangle )\\&=\langle \psi |{\hat {\mathbf {r} }}|\psi \rangle +\mathbf {x} \end{aligned}}}$ using the normalization condition ${\displaystyle \langle \psi |\psi \rangle =1}$, 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.

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

${\displaystyle \langle \mathbf {r} _{T}|{\hat {H}}|\mathbf {r} _{T}\rangle =\langle \mathbf {r} |{\hat {H}}|\mathbf {r} \rangle }$

or

${\displaystyle \langle \mathbf {r} |{\hat {T}}^{-1}{\hat {H}}{\hat {T}}|\mathbf {r} \rangle =\langle \mathbf {r} |{\hat {H}}|\mathbf {r} \rangle }$

This implies that

${\displaystyle {\hat {T}}^{-1}{\hat {H}}{\hat {T}}={\hat {H}}}$

${\displaystyle {\hat {H}}{\hat {T}}-{\hat {T}}{\hat {H}}=[{\hat {H}},{\hat {T}}]=0}$

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).

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

${\displaystyle V(\mathbf {r+R} )=V(\mathbf {r} )}$

for all Bravais lattice vectors ${\displaystyle \mathbf {R} }$

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[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 ${\displaystyle \mathbf {R} }$:

${\displaystyle {\hat {T}}_{\mathbf {R} }f(\mathbf {r} )=f(\mathbf {r} +\mathbf {R} )}$

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.

${\displaystyle {\hat {T}}_{\mathbf {R} _{1}}{\hat {T}}_{\mathbf {R} _{2}}={\hat {T}}_{\mathbf {R} _{2}}{\hat {T}}_{\mathbf {R} _{1}}={\hat {T}}_{\mathbf {R_{1}+R_{2}} }}$

In addition, as the Hamiltonian is periodic, we have,

${\displaystyle {\hat {T}}_{\mathbf {R} }{\hat {H}}={\hat {H}}{\hat {T}}_{\mathbf {R} }}$

Hence, the ${\displaystyle {\hat {T}}_{\mathbf {R} }}$ for all Bravais lattice vectors ${\displaystyle \mathbf {R} }$ and the Hamiltonian ${\displaystyle {\hat {H}}}$ form a set of commutating operators. Therefore, the eigenstates of ${\displaystyle {\hat {H}}}$ can be chosen to be simultaneous eigenstates of all the ${\displaystyle {\hat {T}}_{\mathbf {R} }}$:

${\displaystyle {\hat {H}}\psi ={\mathcal {E}}\psi }$

${\displaystyle {\hat {T}}_{\mathbf {R} }\psi =c(\mathbf {R} )\psi }$

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

${\displaystyle {\hat {T}}_{\mathbf {R} _{1}}{\hat {T}}_{\mathbf {R} _{2}}={\hat {T}}_{\mathbf {R} _{2}}{\hat {T}}_{\mathbf {R} _{1}}={\hat {T}}_{\mathbf {R_{1}+R_{2}} }}$

We have,

${\displaystyle {\begin{aligned}{\hat {T}}_{\mathbf {R} _{1}}{\hat {T}}_{\mathbf {R} _{2}}\psi &=c(\mathbf {R} _{1}){\hat {T}}_{\mathbf {R} _{2}}\psi \\&=c(\mathbf {R} _{1})c(\mathbf {R} _{2})\psi \end{aligned}}}$

And,

${\displaystyle {\hat {T}}_{\mathbf {R_{1}+R_{2}} }\psi =c(\mathbf {R} _{1}+\mathbf {R} _{2})\psi }$

Therefore, it follows that,

${\displaystyle c(\mathbf {R} _{1}+\mathbf {R} _{2})=c(\mathbf {R} _{1})c(\mathbf {R} _{2})}$

Now let the ${\displaystyle \mathbf {a} _{i}}$'s be the three primitive vector for the Bravais lattice. By a suitable choice of ${\displaystyle x_{i}}$, we can always write ${\displaystyle c(\mathbf {a} _{i})}$ in the form

${\displaystyle c(\mathbf {a} _{i})=e^{2\pi ix_{i}}}$

If ${\displaystyle \mathbf {R} }$ is a general Bravais lattice vector, given by

${\displaystyle \mathbf {R} =n_{1}\mathbf {a} _{1}+n_{2}\mathbf {a} _{2}+n_{3}\mathbf {a} _{3}}$

it follows then,

${\displaystyle {\begin{aligned}c(\mathbf {R} )&=c(n_{1}\mathbf {a} _{1}+n_{2}\mathbf {a} _{2}+n_{3}\mathbf {a} _{3})\\&=c(n_{1}\mathbf {a} _{1})c(n_{2}\mathbf {a} _{2})c(n_{3}\mathbf {a} _{3})\\&=c(\mathbf {a} _{1})^{n_{1}}c(\mathbf {a} _{2})^{n_{2}}c(\mathbf {a} _{3})^{n_{3}}\end{aligned}}}$

Substituting ${\displaystyle c(\mathbf {a} _{i})=e^{2\pi ix_{i}}}$ one gets,

${\displaystyle {\begin{aligned}c(\mathbf {R} )&=e^{2\pi i(n_{1}x_{1}+n_{2}x_{2}+n_{3}x_{3})}\\&=e^{i\mathbf {k} \cdot \mathbf {R} }\end{aligned}}}$

where ${\displaystyle \mathbf {k} =x_{1}\mathbf {b} _{1}+x_{2}\mathbf {b} _{2}+x_{3}\mathbf {b} _{3}}$ and the ${\displaystyle \mathbf {b} _{i}}$'s are the reciprocal lattice vectors satisfying the equation ${\displaystyle \mathbf {b} _{i}\cdot \mathbf {a} _{j}=2\pi \delta _{ij}}$

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 ${\displaystyle \mathbf {R} }$,

${\displaystyle {\begin{aligned}\psi (\mathbf {r+R} )&={\hat {T}}_{\mathbf {R} }\psi (\mathbf {r} )\\&=c(\mathbf {R} )\psi (\mathbf {r} )\\&=e^{i\mathbf {k} \cdot \mathbf {R} }\psi (\mathbf {r} )\end{aligned}}}$

So,

This result is known as Bloch's Theorem.

Translational Invariance: Time evolution of the wave functions.

In the passive transformation picture, translational invariance requires,

${\displaystyle [{\hat {T}}(\mathbf {x} ),{\hat {H}}]=0}$

It follows that

${\displaystyle [{\hat {T}}(\mathbf {x} ),{\hat {U}}(t)]=0}$

where ${\displaystyle {\hat {U}}(t)}$ is the unitary time evolution operator.[8] When the Hamiltonian is time independent,

${\displaystyle {\hat {U}}(t)=\exp \left({\frac {-i{\hat {H}}t}{\hbar }}\right)}$

If the Hamiltonian is time dependent, the above commutation relation is satisfied if ${\displaystyle {\hat {p}}}$ or ${\displaystyle {\hat {T}}(\mathbf {x} )}$ commutes with ${\displaystyle {\hat {H}}(t)}$ for all t.

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