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‎doc/pub/week4/html/week4-bs.html‎

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‎doc/pub/week4/html/week4-reveal.html‎

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@@ -1896,6 +1896,348 @@ <h2 id="and-without-the-noise-model">And without the noise model </h2>
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</div>
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</section>
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<section>
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<h2 id="hamiltonians-one-qubit-system">Hamiltonians, one qubit system </h2>
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<p>We start with a simple \( 2\times 2 \) Hamiltonian matrix expressed in
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terms of Pauli \( X \) and \( Z \) matrices, as discussed in the project text as well.
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</p>
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<p>We define a symmetric matrix \( H\in {\mathbb{R}}^{2\times 2} \)</p>
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<p>&nbsp;<br>
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$$
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H = \begin{bmatrix} H_{11} & H_{12} \\ H_{21} & H_{22}
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\end{bmatrix},
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="rewriting-the-hamiltonian">Rewriting the Hamiltonian </h2>
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<p>We let \( H = H_0 + H_I \), where</p>
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<p>&nbsp;<br>
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$$
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H_0= \begin{bmatrix} E_1 & 0 \\ 0 & E_2\end{bmatrix},
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$$
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<p>&nbsp;<br>
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<p>is a diagonal matrix. Similarly,</p>
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<p>&nbsp;<br>
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$$
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H_I= \begin{bmatrix} V_{11} & V_{12} \\ V_{21} & V_{22}\end{bmatrix},
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$$
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<p>&nbsp;<br>
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<p>where \( V_{ij} \) represent various interaction matrix elements.
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We can view \( H_0 \) as the non-interacting solution
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</p>
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<p>&nbsp;<br>
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$$
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\begin{equation}
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H_0\vert 0 \rangle =E_1\vert 0 \rangle,
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\tag{2}
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\end{equation}
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$$
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<p>&nbsp;<br>
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<p>and</p>
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<p>&nbsp;<br>
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$$
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\begin{equation}
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H_0\vert 1\rangle =E_2\vert 1\rangle,
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\tag{3}
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\end{equation}
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$$
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<p>&nbsp;<br>
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<p>where we have defined the orthogonal computational one-qubit basis states \( \vert 0\rangle \) and \( \vert 1\rangle \).</p>
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</section>
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<section>
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<h2 id="using-pauli-matrices">Using Pauli matrices </h2>
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<p>We rewrite \( H \) (and \( H_0 \) and \( H_I \)) via Pauli matrices</p>
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<p>&nbsp;<br>
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$$
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H_0 = \mathcal{E} I + \Omega \sigma_z, \quad \mathcal{E} = \frac{E_1
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+ E_2}{2}, \; \Omega = \frac{E_1-E_2}{2},
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$$
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<p>&nbsp;<br>
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<p>and</p>
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<p>&nbsp;<br>
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$$
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H_I = c \boldsymbol{I} +\omega_z\sigma_z + \omega_x\sigma_x,
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$$
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<p>&nbsp;<br>
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<p>with \( c = (V_{11}+V_{22})/2 \), \( \omega_z = (V_{11}-V_{22})/2 \) and \( \omega_x = V_{12}=V_{21} \).
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We let our Hamiltonian depend linearly on a strength parameter \( \lambda \)
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</p>
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<p>&nbsp;<br>
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$$
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H=H_0+\lambda H_\mathrm{I},
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$$
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<p>&nbsp;<br>
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<p>with \( \lambda \in [0,1] \), where the limits \( \lambda=0 \) and \( \lambda=1 \)
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represent the non-interacting (or unperturbed) and fully interacting
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system, respectively. The model is an eigenvalue problem with only
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two available states.
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</p>
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</section>
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<section>
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<h2 id="selecting-parameters">Selecting parameters </h2>
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<p>Here we set the parameters \( E_1=0 \),
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\( E_2=4 \), \( V_{11}=-V_{22}=3 \) and \( V_{12}=V_{21}=0.2 \).
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</p>
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<p>The non-interacting solutions represent our computational basis.
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Pertinent to our choice of parameters, is that at \( \lambda\geq 2/3 \),
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the lowest eigenstate is dominated by \( \vert 1\rangle \) while the upper
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is \( \vert 0 \rangle \). At \( \lambda=1 \) the \( \vert 0 \rangle \) mixing of
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the lowest eigenvalue is \( 1\% \) while for \( \lambda\leq 2/3 \) we have a
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\( \vert 0 \rangle \) component of more than \( 90\% \). The character of the
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eigenvectors has therefore been interchanged when passing \( z=2/3 \). The
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value of the parameter \( V_{12} \) represents the strength of the coupling
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between the two states.
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</p>
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</section>
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<section>
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<h2 id="setting-up-the-matrix">Setting up the matrix </h2>
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<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="inner_cell">
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<div class="input_area">
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<div class="highlight" style="background: #eeeedd">
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<pre style="font-size: 80%; line-height: 125%;"><span style="color: #8B008B; font-weight: bold">from</span> <span style="color: #008b45; text-decoration: underline">matplotlib</span> <span style="color: #8B008B; font-weight: bold">import</span> pyplot <span style="color: #8B008B; font-weight: bold">as</span> plt
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<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">numpy</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">np</span>
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dim = <span style="color: #B452CD">2</span>
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Hamiltonian = np.zeros((dim,dim))
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e0 = <span style="color: #B452CD">0.0</span>
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e1 = <span style="color: #B452CD">4.0</span>
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Xnondiag = <span style="color: #B452CD">0.20</span>
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Xdiag = <span style="color: #B452CD">3.0</span>
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Eigenvalue = np.zeros(dim)
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<span style="color: #228B22"># setting up the Hamiltonian</span>
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Hamiltonian[<span style="color: #B452CD">0</span>,<span style="color: #B452CD">0</span>] = Xdiag+e0
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Hamiltonian[<span style="color: #B452CD">0</span>,<span style="color: #B452CD">1</span>] = Xnondiag
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Hamiltonian[<span style="color: #B452CD">1</span>,<span style="color: #B452CD">0</span>] = Hamiltonian[<span style="color: #B452CD">0</span>,<span style="color: #B452CD">1</span>]
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Hamiltonian[<span style="color: #B452CD">1</span>,<span style="color: #B452CD">1</span>] = e1-Xdiag
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<span style="color: #228B22"># diagonalize and obtain eigenvalues, not necessarily sorted</span>
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EigValues, EigVectors = np.linalg.eig(Hamiltonian)
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permute = EigValues.argsort()
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EigValues = EigValues[permute]
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<span style="color: #228B22"># print only the lowest eigenvalue</span>
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<span style="color: #658b00">print</span>(EigValues[<span style="color: #B452CD">0</span>])
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</pre>
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</div>
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</div>
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</div>
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</div>
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<div class="output_wrapper">
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<div class="output">
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<div class="output_area">
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<div class="output_subarea output_stream output_stdout output_text">
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</div>
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</div>
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</div>
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</div>
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</div>
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<p>Now rewrite it in terms of the identity matrix and the Pauli matrix X and Z</p>
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<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
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<div class="cell border-box-sizing code_cell rendered">
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<div class="highlight" style="background: #eeeedd">
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<pre style="font-size: 80%; line-height: 125%;"><span style="color: #228B22"># Now rewrite it in terms of the identity matrix and the Pauli matrix X and Z</span>
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X = np.array([[<span style="color: #B452CD">0</span>,<span style="color: #B452CD">1</span>],[<span style="color: #B452CD">1</span>,<span style="color: #B452CD">0</span>]])
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Y = np.array([[<span style="color: #B452CD">0</span>,-<span style="color: #B452CD">1</span>j],[<span style="color: #B452CD">1</span>j,<span style="color: #B452CD">0</span>]])
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Z = np.array([[<span style="color: #B452CD">1</span>,<span style="color: #B452CD">0</span>],[<span style="color: #B452CD">0</span>,-<span style="color: #B452CD">1</span>]])
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<span style="color: #228B22"># identity matrix</span>
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I = np.array([[<span style="color: #B452CD">1</span>,<span style="color: #B452CD">0</span>],[<span style="color: #B452CD">0</span>,<span style="color: #B452CD">1</span>]])
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epsilon = (e0+e1)*<span style="color: #B452CD">0.5</span>; omega = (e0-e1)*<span style="color: #B452CD">0.5</span>
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c = <span style="color: #B452CD">0.0</span>; omega_z=Xdiag; omega_x = Xnondiag
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Hamiltonian = (epsilon+c)*I+(omega_z+omega)*Z+omega_x*X
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EigValues, EigVectors = np.linalg.eig(Hamiltonian)
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permute = EigValues.argsort()
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EigValues = EigValues[permute]
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<span style="color: #228B22"># print only the lowest eigenvalue</span>
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<span style="color: #658b00">print</span>(EigValues[<span style="color: #B452CD">0</span>])
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</pre>
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</div>
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</div>
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</div>
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</div>
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<div class="output_wrapper">
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<div class="output">
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<div class="output_area">
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<div class="output_subarea output_stream output_stdout output_text">
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</div>
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</div>
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</div>
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</div>
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</div>
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</section>
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<section>
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<h2 id="where-are-we-going">Where are we going? </h2>
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<p>For a one-qubit system we can reach every point on the Bloch sphere
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(as discussed earlier) with a rotation about the \( x \)-axis and the
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\( y \)-axis.
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</p>
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<p>We can express this mathematically through the following operations (see whiteboard for the drawing), giving us a new state \( \vert \psi\rangle \)</p>
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<p>&nbsp;<br>
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$$
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\vert\psi\rangle = R_y(\phi)R_x(\theta)\vert 0 \rangle.
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="possible-ansatzes">Possible ansatzes </h2>
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<p>We can produce multiple ansatzes for the new state in terms of the
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angles \( \theta \) and \( \phi \). With these ansatzes we can in turn
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calculate the expectation value of the above Hamiltonian, now
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rewritten in terms of various Pauli matrices (and thereby gates), that is compute
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</p>
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<p>&nbsp;<br>
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$$
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\langle \psi \vert (c+\mathcal{E})\boldsymbol{I} + (\Omega+\omega_z)\boldsymbol{\sigma}_z + \omega_x\boldsymbol{\sigma}_x\vert \psi \rangle.
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$$
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<p>&nbsp;<br>
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<p>We can now set up a series of ansatzes for \( \vert \psi \rangle \) as
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function of the angles \( \theta \) and \( \phi \) and find thereafter the
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variational minimum using for example a gradient descent method.
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</p>
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</section>
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<section>
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<h2 id="more-on-rotation-operators">More on rotation operators </h2>
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<p>To do so, we need to remind ourselves about the mathematical expressions for
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the rotational matrices/operators.
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</p>
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<p>&nbsp;<br>
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$$
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R_x(\theta)=\cos{\frac{\theta}{2}}\boldsymbol{I}-\imath \sin{\frac{\theta}{2}}\boldsymbol{\sigma}_x,
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$$
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<p>&nbsp;<br>
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<p>and</p>
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<p>&nbsp;<br>
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$$
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R_y(\phi)=\cos{\frac{\phi}{2}}\boldsymbol{I}-\imath \sin{\frac{\phi}{2}}\boldsymbol{\sigma}_y.
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="code-example">Code example </h2>
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<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="inner_cell">
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<div class="input_area">
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<div class="highlight" style="background: #eeeedd">
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<pre style="font-size: 80%; line-height: 125%;"><span style="color: #228B22"># define the rotation matrices</span>
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<span style="color: #228B22"># Define angles theta and phi</span>
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theta = <span style="color: #B452CD">0.5</span>*np.pi; phi = <span style="color: #B452CD">0.2</span>*np.pi
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Rx = np.cos(theta*<span style="color: #B452CD">0.5</span>)*I-<span style="color: #B452CD">1</span>j*np.sin(theta*<span style="color: #B452CD">0.5</span>)*X
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Ry = np.cos(phi*<span style="color: #B452CD">0.5</span>)*I-<span style="color: #B452CD">1</span>j*np.sin(phi*<span style="color: #B452CD">0.5</span>)*Y
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<span style="color: #228B22">#define basis states</span>
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basis0 = np.array([<span style="color: #B452CD">1</span>,<span style="color: #B452CD">0</span>])
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basis1 = np.array([<span style="color: #B452CD">0</span>,<span style="color: #B452CD">1</span>])
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NewBasis = Ry @ Rx @ basis0
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<span style="color: #658b00">print</span>(NewBasis)
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<span style="color: #228B22"># Compute the expectation value</span>
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<span style="color: #228B22">#Note hermitian conjugation</span>
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Energy = NewBasis.conj().T @ Hamiltonian @ NewBasis
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<span style="color: #658b00">print</span>(Energy)
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</pre>
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</div>
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</div>
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</div>
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</div>
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<div class="output_wrapper">
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<div class="output">
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</div>
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</div>
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</div>
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</div>
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</div>
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<p>Not an impressive results. We set up now a loop over many angles \( \theta \) and \( \phi \) and compute the energies</p>
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<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="highlight" style="background: #eeeedd">
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<pre style="font-size: 80%; line-height: 125%;"><span style="color: #228B22"># define a number of angles</span>
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n = <span style="color: #B452CD">20</span>
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angle = np.arange(<span style="color: #B452CD">0</span>,<span style="color: #B452CD">180</span>,<span style="color: #B452CD">10</span>)
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n = np.size(angle)
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ExpectationValues = np.zeros((n,n))
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<span style="color: #8B008B; font-weight: bold">for</span> i <span style="color: #8B008B">in</span> <span style="color: #658b00">range</span> (n):
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theta = np.pi*angle[i]/<span style="color: #B452CD">180.0</span>
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Rx = np.cos(theta*<span style="color: #B452CD">0.5</span>)*I-<span style="color: #B452CD">1</span>j*np.sin(theta*<span style="color: #B452CD">0.5</span>)*X
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<span style="color: #8B008B; font-weight: bold">for</span> j <span style="color: #8B008B">in</span> <span style="color: #658b00">range</span> (n):
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phi = np.pi*angle[j]/<span style="color: #B452CD">180.0</span>
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Ry = np.cos(phi*<span style="color: #B452CD">0.5</span>)*I-<span style="color: #B452CD">1</span>j*np.sin(phi*<span style="color: #B452CD">0.5</span>)*Y
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NewBasis = Ry @ Rx @ basis0
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Energy = NewBasis.conj().T @ Hamiltonian @ NewBasis
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Edifference=<span style="color: #658b00">abs</span>(np.real(EigValues[<span style="color: #B452CD">0</span>]-Energy))
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ExpectationValues[i,j]=Edifference
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<span style="color: #658b00">print</span>(np.min(ExpectationValues))
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</pre>
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</div>
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</div>
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</div>
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</div>
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<div class="output_wrapper">
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<div class="output">
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<div class="output_subarea output_stream output_stdout output_text">
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</div>
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</div>
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</div>
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</div>
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</div>
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<p>Clearly, this is not the very best way of proceeding. Rather, here we
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would compute the gradient and thereby find the minimum as function of
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the angles \( \theta \) and \( \phi \). This will lead us to the so-called Variational Quantum Eigensolver to be discussed next week.
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</p>
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</section>
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18992241

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</div> <!-- class="slides" -->

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