# Spatial translations from Wigner's theorem

Some books argue that typical coordinate transformations such as space translations and rotations are represented in quantum mechanics by unitary operators because the Wigner's theorem. However I do not find any clear proof of this. For instance, suppose 1D for the sake of simplicity, by definition spatial translations change the position operator as $$\hat{x}\to \hat{x}'=\hat{x}+a$$ where $a$ is a constant.

Why this transformation satisfies the premises of the Wigner's theorem? Namely, bijectivity and preservation of absolute value of scalar products $|\langle \psi_1|\psi_2\rangle|=|\mbox{}'\langle \psi_1|\psi_2\rangle'|$.

More specifically, if $|x\rangle$ is a position eigenstate $\hat{x}|x\rangle=x|x\rangle$, then $|x\rangle$ is obviously an eigenstate of $\hat{x}'$ with eigenvalue $(x+a)$. Therefore a spatial translation change a position eigenstate as $$|x\rangle\to |x\rangle'=|x+a\rangle.$$ However to check bijectivity and the condition on scalar products we need to obtain the transformation law for any state $$|\psi\rangle=\int_{-\infty}^{\infty}\psi(x)|x\rangle.$$ Under linearity this state changes to $$|\psi\rangle'=\int_{-\infty}^{\infty}\psi(x)|x\rangle'=\int_{-\infty}^{\infty}\psi(x)|x+a\rangle,$$ but we cannot assume linearity because we actually want to obtain it as a result of applying the Wigner's theorem.

• Be so kind and remind us what the Wigner theorem states please. – flippiefanus Dec 28 '17 at 3:25
• In short: Wigner's theorem states that any bijective map between state vectors $|\psi\rangle\to|\psi\rangle'$ such that $|\langle\psi_1|\psi_2\rangle|=|\mbox{}'\langle\psi_1|\psi_2\rangle'|$ can be implemented by an operator $T$, $|\psi\rangle'=T|\psi\rangle$, which is either linear and unitary, or antilinear and antiunitary. – NessunDorma Dec 28 '17 at 9:40
• Wigner's theorem is a statement about symmetry transformations, i.e. mappings between (unit) rays in Hilbert space. (Anti)linearity is proven when this mapping is "lifted" from $\mathscr{P} \mathcal{H}$ to $\mathcal H$, being required by the well-definedness of the operator acting on vectors – DanielC Dec 28 '17 at 16:10
• @NessunDorma. Your post is not fully clear. You mention the symmetry operation acting on the position observable (which the Wigner's theorem does not address), then later on the state vector (which is the subject of Wigner's theorem). Pay attention that position eigenstates are not normalizable, thus you need to extend (in a particular way) Wigner's theorem to rigged Hilbert spaces to obtain the correct picture for continuous coordinates. – DanielC Dec 28 '17 at 16:15
• @DanielC, I'm interested in proving that a transformation such as a space translation in quantum mechanics is a symmetry transformation. In classical mechanics a space translation is a change in the position $x\to x+a$. In quantum mechanics such a change can be formulated analogously with operators $\hat{x}\to \hat{x}+a$. Now the question is why such a change can be implemented by a unitary operation? Some books say this is because the Wigner's theorem, i.e. because it is a symmetry transformation (an isometric one-to-one correspondence between unit rays). How to prove this? – NessunDorma Dec 28 '17 at 17:05