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I have a confusion with regards to the principle of QM that states that time evolution must be unitary. In particular, given that states transform through time as $|\Psi(t)\rangle = U(t)|\Psi(0)\rangle$; does the condition: $$\langle\Phi(0)|\Psi(0)\rangle=0 \ \implies \ \langle\Phi(t)|\Psi(t)\rangle=0 $$

imply that $U$ must be unitary, or is it imposed on $U$?

I realize that with the before mentioned condition it is posible to prove that for two orthonormal memebers of the basis, $i,j$: $\langle i| U^{\dagger}U | j \rangle = 0$. Is it posible to prove that for the same member the product is 1?

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  • $\begingroup$ If $U$ is unitary then $U^{\dagger}$ is the inverse of $U$ by definition. $\endgroup$ – ZeroTheHero Sep 27 '17 at 14:10
  • $\begingroup$ Yes, but the question is about how to show that, that is the case $\endgroup$ – N. Bitar Sep 27 '17 at 14:18
  • $\begingroup$ Since $\vert \Psi(t)\rangle = U\vert\Psi(0)\rangle$ then clearly $$ \langle \Phi(t)\vert \Psi(t)\rangle = \langle \Phi(0)\vert U^\dagger\,U\vert \Psi(0)\rangle = \langle\Phi(0)\vert\Psi(0)\rangle $$ valid $\forall t$ implies $U^\dagger U=\hat 1\, \forall t$ as well, provided your kets are arbitrary. (Hopefully this should be enough.) $\endgroup$ – ZeroTheHero Sep 27 '17 at 14:32
  • $\begingroup$ The condition of orthogonality preservation you stated alone would also allow for operators that satisfy $U^\dagger U = c \cdot 1$ with any constant $c$. Usually one says that unitarity comes from norm preservation, i.e. $< \psi(t) ,\psi(t)> = <\psi(0),\psi(0)>$. $\endgroup$ – Luke Sep 27 '17 at 15:23
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It is an interesting elementary problem. After I while I proved the following proposition where I use $A^*$ for the adjoint of $A$.

Proposition. Let $U: H \to H$ be a bounded operator over a Hilbert space $H$. The following conditions are equivalent.

(a) For $x,y \in H$,$\quad$ $\langle x|y\rangle =0$ implies $\langle Ux|Uy\rangle =0$

(b) $U^*U = cI$ for some real $c\geq 0$.

Before proving the statement I observe that even if $c=1$, $U$ is not necessarily unitary, because unitarity is $U^*U=UU^*=I$. And here $UU^*=I$ generally fails when $H$ is infinite dimensional (otherwise it is trivially true as a consequence of $U^*U=I$). For $c=1$, $U$ is an isometry not necessarily surjective.

Proof. It is obvious that (b) implies (a), so we prove that (a) implies (b). Condition (a) can be rephrased as $y \perp x$ implies $y \perp U^*Ux$. As a consequence $U^*Ux \in \{\{x\}^\perp\}^\perp$ which is the linear span of $x$. In other words $U^*U x = \lambda_x x$ for some $\lambda_x\in \mathbb C$. My goal is now proving that $\lambda_x$ does not depend on $x$.

To this end, consider a couple of vectors $x \perp y$ with $x, y \neq 0$. Using the argument above we have $$U^*U x = \lambda_x x\:,\quad U^*U y = \lambda_y y\:, \quad U^*U (x+y) = \lambda_{x+y} (x+y)\:.\tag{1}$$ Linearity of $U^*U$ applied to the last identity leads to $$U^*Ux + U^*Uy = \lambda_{x+y}x + \lambda_{x+y}y\:,$$ namely $$U^*Ux- \lambda_{x+y}x = -(U^*Uy- \lambda_{x+y}y)\:\:.$$ Exploiting the first two identities in (1) we get $$(\lambda_x- \lambda_{x+y})x = -(\lambda_y- \lambda_{x+y})y\:.$$ Since $x \perp y$ and $x,y \neq 0$, the only possibility is that $$\lambda_x = \lambda_{x+y} = \lambda_y\:.$$ So a couple of orthogonal non-vanishing vectors has the same $\lambda_x$.

To conclude consider a Hilbert basis $\{x_n\}$ of $H$ so that, if $z\in H$, $$z = \sum_n c_n x_n \tag{2}$$ for complex numbers $c_n$. Since $U^*U$ is continuous ($U$ is bounded), $$U^*Uz = \sum_n c_n U^*Ux_n = \sum_n c_n \lambda_{x_n}x_n\tag{3}$$ But we know from the previous argument that $\lambda_{x_n} = \lambda_{x_m}$ so that, indicating with $c$ the common value of the $\lambda_{x_n}$, (3) can be rewritten as $$U^*Uz = \sum_n c_n cx_n = c\sum_n c_n x_n = cz\:.$$ Since $z\in H$ was arbitrary, we have found that $$U^*U=c I\:.$$ Taking the adjoint of both sides we obtain $c=\overline{c}$ so that $c$ is real. Finally, $$0 \leq \langle Ux | Ux\rangle = \langle x| U^*U x\rangle = c \langle x| x \rangle$$ so that $c\geq 0$. QED

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  • $\begingroup$ This is nice but does the part "Is it posible to prove that for the same member the product is 1?" not imply $c=\lambda_x=1$? $\endgroup$ – ZeroTheHero Sep 27 '17 at 19:32
  • $\begingroup$ I do not understand, we already know that $U$ such thst $U^*U= cI$ satisfies the initial hypotheses also for $c\neq 1$. If it were possible to prove that $c=1$ we would exclude this case that we know exists. $\endgroup$ – Valter Moretti Sep 27 '17 at 19:40
  • $\begingroup$ Something's clearly throwing me off here. I'll come bac k to in a couple of days. $\endgroup$ – ZeroTheHero Sep 27 '17 at 19:57

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