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Let $H = P$ be the Hamiltonian described by a Pauli operator $P$. The real time evolution according to H is $$e^{-iPt}.$$ While the imaginary time evolution is $$e^{-Pt}.$$

Consider the case $P=Y$. Than we have the y-axis rotation given by $$e^{-iYt},$$ which should be unitary. However, it is also true that $$Y = \begin{pmatrix}0 & -i\\i & 0\end{pmatrix} = -i\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}.$$

If I substitute this matrix into the rotation operator I get $$e^{-\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}}t$$

Isn't this an imaginary time evolution? But it can't be either a unitary and an ITE. What am I doing wrong?

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  • $\begingroup$ Have you checked the eigenvalues of the matrix? $\endgroup$
    – TLDR
    Commented Dec 6 at 23:56
  • $\begingroup$ I guess they are +i and -i, the same of Y. Why? What does this mean? $\endgroup$
    – Daniele Cuomo
    Commented Dec 7 at 0:10
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    $\begingroup$ Real (unitary) time evolution doesn't necessarily imply that the matrix elements of the propagator $e^{iHt}$ (or the wave functions/vectors that it operates on) have a non-zero imaginary part. Since the eigenvalues of $i\sigma_Yt$ are purely imaginary, and the eigenvectors of $\sigma_Yt$ form an orthonormal basis, $e^{i\sigma_Y t}$ preserves the complex norm of the amplitudes of its eigenvectors, and encodes dynamics in their relative phases; this is the defining feature of unitary (von Neumann-entropy preserving) time evolution (imaginary time evolution does not preserve the complex norm.) $\endgroup$
    – TLDR
    Commented Dec 7 at 0:29
  • $\begingroup$ I see. May you explain this step by step as an answer? $\endgroup$
    – Daniele Cuomo
    Commented Dec 7 at 0:39
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    $\begingroup$ @DanieleCuomo - Eigenvalues of $Y$ are $\pm1$, not $\pm i$. $\endgroup$
    – Prahar
    Commented Dec 7 at 10:22

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The eigenvalues of the Pauli matrix $$\sigma_y= \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right)$$ are $+1$ and $-1$ with associated (normalized) eigenvectors $$\chi_+= \frac{1}{\sqrt{2}} \left( \begin{array} {c} 1 \\ i \end{array} \right), \quad \chi_-=\frac{1}{\sqrt{2}}\left(\begin{array} {c} 1\\ -i\end{array} \right). $$ The spectral representation of $\sigma_y$ is thus given by $$\sigma_y = \chi_+ \chi_+^\dagger -\chi_- \chi_-^\dagger,$$ where $\chi_+\chi_+^\dagger$ and $\chi_-\chi_-^\dagger$ are the orthogonal projection operators on the two eigenspaces. An arbitrary function $f$ of $\sigma_y$ takes the form $$f(\sigma_y)=f(1)\chi_+\chi_+^\dagger+f(-1)\chi_-\chi_-^\dagger.$$ For the special case of the unitary operator$f(\sigma_y)=\exp(-i\sigma_y t)$ one finds $$e^{-i \sigma_y t}=e^{-it} \chi_+\chi_+^\dagger + e^{it}\chi_-\chi_-^\dagger=\frac{e^{-it}}{2}\left(\begin{array} {cc} 1 &-i \\i & 1 \end{array} \right)+\frac{e^{it}}{2}\left( \begin{array} {cc}1 & i \\ -i &1\end{array} \right) =\left( \begin{array} {cc} \cos t & -\sin t\\ \sin t & \cos t \end{array} \right) . $$

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