# Evolution operator for “blending” a pair of eigenstates

Consider a measurement operator ("observable") $\hat O$ which has ("a spectrum of") only two distinct eigenvectors; formally

$$\hat O |\bullet\rangle := r_{\bullet}~|\bullet\rangle, \qquad \hat O |\circ\rangle := r_{\circ}~|\circ\rangle, \qquad \langle \bullet |\circ\rangle = 0,$$

where the eigenvalues $r_{\bullet}$ and $r_{\circ}$ are (suitable) real numbers.

Is there a time-dependent unitary operator $\hat U[~t~]$ such that

$$\frac{|\bullet\rangle + |\circ\rangle}{\sqrt 2} = \hat U[~t_{\text{blend}}~] ~ |\bullet\rangle$$ ?

If so, can this operator be expressed in the form $\hat U[~t~] := \text{Exp}[~\frac{-i~t}{\hbar}~\hat H[~t~]~]$ ?

If so, can the corresponding operator $\hat H[~t~]$ be expressed in terms of the given operator $\hat O$, or its eigenvectors or eigenvalues, and, as necessary, the coordinate value $t_{\text{blend}}$ and the coordinate variable $t$?

Edit
(along with correcting the above "evolution equation" stated within the bra-ket formalism):

Within the density matrix formalism it could be asked correspondingly whether there is a time-dependent unitary operator $\hat U[~t~]$ such that

$$\sqrt{\frac{1}{2}} \left( \array{ |\bullet\rangle\langle\bullet | & |\bullet\rangle\langle\circ | \cr |\circ\rangle\langle\bullet | & |\circ\rangle\langle\circ | } \right) = \hat U[~t_{\text{blend}}~] ~ \left( \begin{array}{cc} |\bullet\rangle\langle\bullet | & 0 \\ 0 & |\circ\rangle\langle\circ | \end{array} \right) ~ \hat U[~t_{\text{blend}}~]^{\dagger}.$$

OK firstly operators act on kets as vectors, so for some unitary operator $\hat{U}$, the action of $\hat{U}$ sends $|\psi\rangle\rightarrow \hat{U}|\psi\rangle$ not $\hat{U}|\psi\rangle\hat{U}^\dagger$, which is normally not a well defined expression.

For the rest of your question, since we are only dealing with a two dimensional space (or at least only a two dimensional subspace of some larger space) it is clearest to use matrix notation. So $$\hat{U}(t) = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\quad |\bullet\rangle = \left(\begin{array}{c}1\\0\end{array}\right)\quad |\circ\rangle = \left(\begin{array}{c}0\\1\end{array}\right)$$

In this notation your blending condition becomes $$\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1 \end{array}\right) = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right) = \left(\begin{array}{c}a\\c\end{array}\right)$$ The requirement that $\hat{U}$ be unitary then fixes $b$ and $d$ up to a sign, so it is easy to work out that$$\hat{U} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& \pm 1\\1&\mp 1\end{array}\right)$$

If we put the minus sign in the top right (the case where it is in the bottom right differs only by a change of basis) then $\hat{U}$ has a standard representation in terms of Pauli matrices$$\hat{U} = \exp\left(-\imath\frac{\pi}{4}\sigma^{\,y}\right)$$

• By Symmetry: "OK firstly operators act on kets as vectors [...] $|\psi\rangle \rightarrow \hat U |\psi\rangle$" -- Aw; thanks for clarifying (+1). So, what I wrote at least doesn't work with "bra-ket" notation ... (I may need a while for deciding whether and how to correct/edit my OP regarding this "formality".) "[...] clearest to use matrix notation." -- Great. However: I'm missing some explicit $t$ dependence in your expressions for operator $\hat U[~t~]$. (This still seems a substantive part of my question which you answer didn't address.) – user12262 Apr 20 '15 at 5:14
• @user12262 Just replace $\pi/4$ in the expression for $\hat{U}$ by anything of the form $\Omega t$, where $\Omega$ has dimensions of frequency and $t$ is time. Then you get your required unitary operator by setting $\Omega t_\mathrm{blend} = \pi/4$. – Mark Mitchison Apr 20 '15 at 9:59
• @Mark Mitchison: "Just replace $\pi/4$ in the expression for $\hat U$ by anything of the form $\Omega~t$ [with] $\Omega~t_{\text{blend}} = \pi/4$. Can you please prove that: $$\hat U[~t_{\text{blend}}~]=\frac{1}{\sqrt 2}\left(\array{1&-1\cr 1&1}\right) =?= \text{Exp}[~i~\frac{\pi}{4}~\hat \sigma_y~] = \text{Exp}[~i~\frac{\pi}{4}~{\left(\array{0&-i \cr i&0}\right)}~]=\text{Exp}[~\frac{\pi}{4}~\left(\array{ 0 & 1 \cr -1 & 0}\right)~]$$ ? (Meanwhile I ponder this ...) – user12262 Apr 20 '15 at 18:25
• @user12262 firstly I realised I missed a minus sign in my expression for $\hat{U}$. The expression for $\hat{U}$ follows straightforwardly from the power series definition of the matrix exponential and the fact that $\hat{\sigma}_y^2 = \hat{I}$ – By Symmetry Apr 20 '15 at 18:36
• @user12262 You have the identity, which you can prove using the arguments in By Symmetry's comment: $$e^{i \theta\mathbf{x}\cdot\mathbf{\sigma}} = \cos(\theta) + i\hat{\mathbf{x}}\cdot \mathbf{\sigma} \sin(\theta),$$ where $\sigma$ here is a vector of Pauli operators $\sigma^{x,y,z}$, and $\hat{\mathbf{x}}$ is a unit vector parallel to $\mathbf{x}$. – Mark Mitchison Apr 20 '15 at 21:05