I try to solve problems from Problems in Quantum Computing.

I stuck with problem #3:

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I do the following:

Because: $$ \sigma_2 = \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}$$ Then: $$ -i \frac{\phi}{2}\sigma_2 = \begin{pmatrix} 0 & -\frac{\phi}{2}\\ \frac{\phi}{2} & 0 \end{pmatrix} $$

$$\exp\begin{pmatrix} 0 & -\frac{\phi}{2}\\ \frac{\phi}{2} & 0 \end{pmatrix} = \begin{pmatrix} 1 & \exp(-\frac{\phi}{2})\\ \exp(\frac{\phi}{2}) & 1 \end{pmatrix}$$

If I multiply the result of the last calculation with $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ or $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ I can't get $\psi_1(\phi)$ or $\psi_2(\phi)$. I get some unnormalised state like:

$$ \begin{pmatrix} 1 \\ \exp(\frac{\phi}{2}) \end{pmatrix} $$

Does it mean that the definition of the problem is not correct?


Your mistake is computing exponent of matrix; use the formula

$$\exp(i\theta\sigma_2)=\cos(\theta)\cdot I+i\cdot \sin(\theta)\cdot\sigma_2$$

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  • $\begingroup$ I haven't known about this "hack". Could you suggest a site where I can read about it? $\endgroup$ – Kenenbek Arzymatov Dec 6 '19 at 21:12
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    $\begingroup$ This "hack" should be in any good QM or Quantum Information course; see wikipedia; I substituted $\hat{n}=(0,1,0)$ $\endgroup$ – kludg Dec 6 '19 at 21:19
  • $\begingroup$ This is an application of a matrix function definition. See my answer below. $\endgroup$ – Martin Vesely Dec 6 '19 at 23:43

A matrix function $f(A)$ for normal matrix $A$ is defined as follows \begin{equation} f(A)=\sum_{i=1}^{n}f(\lambda_i)v_iv_i^T \end{equation} where $\lambda_{i}$ is an eigenvalue and $v_{i}$ is coresponding eigenvector (note: transposed vector $v_{i}$ is a row vector).

In your case: $f(A) = \mathrm{e}^A$ and $A = -i\frac{\phi}{2}\sigma_{2}$.

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  • $\begingroup$ +1, though a reference stating the formula along with a proof would be appreciated. $\endgroup$ – Sanchayan Dutta Dec 7 '19 at 0:29
  • $\begingroup$ For more information see Nielsen and Chuang, Quantum Computation and Quantum Information, pg. 75 (chapter 2.1.8 - Operator functions) $\endgroup$ – Martin Vesely Dec 7 '19 at 9:07
  • $\begingroup$ @Martin, thanks for your note. One quick question, in this case what is 𝑣𝑖? $\endgroup$ – Parfait Atchadé Jan 16 at 9:25
  • $\begingroup$ please use comments for additional questions. $v_{i}$ is ith eigenvector of matrix $A$. $\endgroup$ – Martin Vesely Jan 16 at 12:16
  • $\begingroup$ Please don't add "thank you" as an answer. Instead, vote up the answers that you find helpful. - From Review $\endgroup$ – met927 Jan 16 at 14:21

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