Translations and momentum

Intuition

Move a state a distance $a$ to the right and its wavefunction slides over: the new value at $x$ is the old value at $x-a$ . Written as a Taylor series, the slide is the exponential of the derivative, and the derivative is the momentum operator in disguise. So momentum is the generator of translations — the deepest reason why it is conserved in empty space, where sliding everything over changes nothing.

Sliding a transparency across a table does not change the picture on it. If the table is featureless, nothing about the picture’s future depends on where it lies; that indifference to place is momentum conservation.

A state and the same state translated by $a$ : the new wavefunction at $x$ is the old one at $x-a$ . The operator that does this is $e^{-ia\hat{p}/\hbar}$ .

The translation operator $\hat{T}(a)$ shifts every state by $a$ . It is unitary, it forms a family with $\hat{T}(a)\hat{T}(b)=\hat{T}(a+b)$ , and its generator is the momentum.

\begin{aligned}\hat{T}(a)\psi(x)&=\psi(x-a)\\ \hat{T}(a)&=e^{-ia\hat{p}/\hbar}\end{aligned}

Properties

$\hat{T}(a)^{\dagger}\hat{x}\hat{T}(a)=\hat{x}+a$ : a translated state has every position average shifted by .

Momentum generates translations

\psi(x-a)=e^{-ia\hat{p}/\hbar}\,\psi(x)

Expand $\psi(x-a)$ in its Taylor series about $x$ . Its terms are the powers of $-a\,d/dx$ divided by factorials, which is the exponential series of that operator. Since $\hat{p} = - i ℏ d /$ , the operator is .

Proof steps

\psi(x-a)=\sum_{n=0}^{\infty}\frac{(-a)^{n}}{n!}\,\psi^{(n)}(x)

Taylor’s series about $x$ .

=\sum_{n=0}^{\infty}\frac{1}{n!}\Bigl(-a\frac{d}{dx}\Bigr)^{n}\psi(x)=e^{-a\,d/dx}\,\psi(x)

The same terms, written as the exponential series of an operator.

\frac{d}{dx}=\frac{i}{\hbar}\hat{p}

From $\hat{p}=-i\hbar\,d/dx$ .

\hat{T}(a)=e^{-ia\hat{p}/\hbar}

Substitute: the translation is the exponential of momentum.

\hat{T}(a)\ket{p}=e^{-ipa/\hbar}\ket{p}

A momentum eigenstate only acquires a phase.

Applications

Practice

Sliding a State

Translating a state by a distance $a$ moves its wavefunction to the right by $a$ : the new wavefunction at $x$ is the old one at $x-a$ .

\hat{T}(a)\psi(x)=\psi(x-a)

Try it

A state has $\psi(x)=e^{-x^{2}}$ . What is its wavefunction after a translation by $+2$ ?