# How does the Hamiltonian changes after rotating the coordinate frame

A basic question in rotation. Initially I have a Hamiltonian $$H = \frac{-1}{2}(4E_c(1-n_g)\sigma_z + E_J\sigma_x) \, .$$ It's said that, if I rotate the coordinate system by mixing angle, $$\theta = \arctan \left(\frac{E_J}{4E_c(1-n_g)} \right)$$ then in the new coordinate system, $\sigma_z$ becomes the quantization angle and the Hamiltonian would be $$H'= \hbar \omega_a \sigma_z$$ where $\hbar \omega_a$ will be $$\hbar \omega_a = \sqrt{E_J^2+(4E_c(1-n_g))^2}$$ Could you please tell how the Hamiltonian changes?

• Is this the Hamiltonian of a charge qubit or something? Jul 19, 2018 at 17:49
• Yes, that's right. Jul 19, 2018 at 18:08
• I have a feeling that it is not a full information. If you want to target a broader audience, give a reference, so that one could figure out such things as rotation group, axis etc. Jul 19, 2018 at 19:13
• Intuitively if try to rotate around axis $\:y\:$ by $\:-\theta\:$ : $U=\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\sigma_{y}$ ??? Jul 19, 2018 at 19:29

## 1 Answer

For convenience in order to simplify the equations let ignore the constant $\:-\frac12\:$ so that we have the Hamiltonian $$H=\mathrm A\sigma_{x}+\mathrm C\sigma_{z}\,,\quad \mathrm A=E_J\,, \:\: \mathrm C=\left[4E_c\left(1-n_g\right)\right] \tag{01}\label{eq1}$$ We must find a rotation in real space $\:\mathbb{R}^{3}\:$ so that for the new Hamiltonian $$H'=\mathrm B\sigma_{z} \tag{02}$$ The rotation must be(1) around the $\:y-$axis through an angle $\:-\theta\:$ represented by the following special unitary matrix
$$\!\!\!\!\!\!U=\cos\frac{\theta}{2}I+i\sin\frac{\theta}{2}\sigma_{y} \,,\quad \theta =\arctan\left(\dfrac{\mathrm A}{\mathrm C}\right)= \arctan \left[\frac{E_J}{4E_c\left(1-n_g\right)} \right]\,,\quad \theta \in [0,\pi] \tag{03}\label{eq3}$$ In the following $\:\omega=\frac{\theta}{2}:$

\begin{align} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!H'&\!\!\!\!\!\!\!\!\!= UHU^{*} =\left(\cos\omega I+i\sin\omega\sigma_{y}\right)\left(\mathrm A\sigma_{x}+\mathrm C\sigma_{z}\right)\left(\cos\omega I-i\sin\omega\sigma_{y}\right) \nonumber\\ & \!\!\!\!\!\!\!\!\!= \left(\mathrm A\cos\omega \sigma_{x}+i\mathrm A\sin\omega\sigma_{y}\sigma_{x}+\mathrm C\cos\omega \sigma_{z}+i\mathrm C\sin\omega\sigma_{y}\sigma_{z}\right)\left(\cos\omega I-i\sin\omega\sigma_{y}\right) \nonumber\\ &\!\!\!\!\!\!\!\!\! = \mathrm A\cos^{2}\omega \sigma_{x}-i\mathrm A\cos\omega\sin\omega\sigma_{x}\sigma_{y}+i\mathrm A\cos\omega\sin\omega\sigma_{y}\sigma_{x}+\mathrm A\sin^{2}\omega\sigma_{y} \sigma_{x}\sigma_{y} \nonumber\\ &\:\!\!\!\!\!\!\!\!\!+\mathrm C\cos^{2}\omega \sigma_{z}-i\mathrm C\cos\omega\sin\omega\sigma_{z}\sigma_{y}+i\mathrm C\cos\omega\sin\omega\sigma_{y}\sigma_{z}+\mathrm C\sin^{2}\omega\sigma_{y} \sigma_{z}\sigma_{y} \nonumber\\ &\!\!\!\!\!\!\!\!\! \!=\!\mathrm A \left(\cos^{2}\omega\!-\!\sin^{2}\omega\right)\sigma_{x}\!+\!2 \mathrm A\cos\omega\sin\omega\sigma_{z}\!-\!2 \mathrm C\cos\omega\sin\omega\sigma_{x}\!+\!\mathrm C \left(\cos^{2}\omega\!-\!\sin^{2}\omega\right)\sigma_{z} \nonumber\\ &\!\!\!\!\!\!\!\!\!\! =\left(\mathrm A\sin\theta+\mathrm C \cos\theta\right)\sigma_{z}+\left(\mathrm A\cos\theta-\mathrm C \sin\theta\right)\sigma_{x}=\left(\mathrm A\sin\theta+\mathrm C \cos\theta\right)\sigma_{z} \tag{04}\label{eq4} \end{align} that is, see equation \eqref{eq11}(2)

$$H'=\sqrt{\mathrm A^{2}+\mathrm C^{2}}\:\sigma_{z}= \sqrt{E_J^{2}+\left[4E_c\left(1-n_g\right)\right]^{2}}\:\sigma_{z} \tag{05}$$

In \eqref{eq4} we make use of Pauli matrices properties, see equations \eqref{eq12a} and \eqref{eq12b}.(3)

(1) The reasoning :

The Pauli matrices $$\sigma_{x}= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \: , \:\:\: \sigma_{y}= \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \: , \:\:\: \sigma_{z}= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{06}$$ are the hermitian traceless matrix representations of the unit vectors $$\mathbf{i}= \begin{bmatrix} 1 \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}} \end{bmatrix} \: , \:\:\: \mathbf{j}= \begin{bmatrix} 0 \vphantom{\tfrac{a}{b}}\\ 1 \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}} \end{bmatrix} \: , \:\:\: \mathbf{k}= \begin{bmatrix} 0 \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}}\\ 1 \vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{07}$$ along the $\:x,y, z\:$ axis respectively according to the following bijection (one-to-one and onto correspondence) between $\mathbb{R}^3$ and the space of $2\times2$ hermitian traceless matrices, let it be $\mathbb{H}$ : $$\mathbf{w}=(x,y,z)\in \mathbb{R}^3\;\longleftarrow\!\longrightarrow \; W= \begin{bmatrix} z & x-iy \\ x+iy & -z \end{bmatrix} \in \mathbb{H} \tag{08}$$ Under this interpretation the Hamiltonian $\:H\:$ of equation \eqref{eq1} is the hermitian traceless matrix representation of the vector $$\!\!\!\!\!\!\mathbf{h}\!=\!\mathrm A \mathbf{i}+\mathrm C \mathbf{k}\!=\! \begin{bmatrix} \mathrm A \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}}\\ \mathrm C \vphantom{\tfrac{a}{b}} \end{bmatrix} \!=\! \begin{bmatrix} E_J \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}}\\ 4E_c\left(1-n_g\right) \vphantom{\tfrac{a}{b}} \end{bmatrix} \, , \: \Vert \mathbf{h}\Vert^{2}=\mathrm A^{2}+\mathrm C^{2}=E_J^{2}+\left[4E_c\left(1-n_g\right)\right]^{2} \tag{09}\label{eq9}$$ But if we want under a rotation the new Hamiltonian to be $\:H'=\mathrm B\sigma_{z}\:$ then this Hamiltonian will be the hermitian traceless matrix representation of the vector $$\mathbf{h'}\!=\!\Vert \mathbf{h}\Vert\:\mathbf{k}\!=\!\sqrt{\mathrm A^{2}+\mathrm C^{2}}\:\mathbf{k}\!=\! \begin{bmatrix} 0 \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}}\\ \sqrt{\mathrm A^{2}+\mathrm C^{2}} \end{bmatrix} = \begin{bmatrix} 0 \vphantom{\tfrac{a}{b}}\\ 0 \vphantom{\tfrac{a}{b}}\\ \sqrt{E_J^{2}+\left[4E_c\left(1-n_g\right)\right]^{2}} \end{bmatrix} \tag{10}\label{eq10}$$ So, the rotation we try to find must rotate the vector $\:\mathbf{h}\:$ of \eqref{eq9} to the vector $\:\mathbf{h'}\:$ of \eqref{eq10}. The most simple rotation we have to do it is that given by \eqref{eq3}.

In the Figure below we see the vectorial in $\:\mathbb{R}^{3}\:$ representation of the $2\times2$ hermitian traceless matrices $\:\sigma_{x},\sigma_{y},\sigma_{z},H,H'$.

(2) $$\dfrac{\sin\theta}{\cos\theta}=\tan\theta \stackrel{defin. (03)}{=\!=\!=\!=\!=\!=}\dfrac{\mathrm A }{\mathrm C }\Longrightarrow \left. \begin{cases} \sin\theta=\dfrac{\mathrm A}{\sqrt{\mathrm A^{2}+\mathrm C^{2}}}\\ \cos\theta=\dfrac{\mathrm C}{\sqrt{\mathrm A^{2}+\mathrm C^{2}}} \end{cases} \right\} \Longrightarrow \left. \begin{cases} \mathrm A\sin\theta+\mathrm C \cos\theta=\sqrt{\mathrm A^{2}+\mathrm C^{2}}\vphantom{\dfrac{\mathrm A}{\sqrt{\mathrm A^{2}+\mathrm C^{2}}}}\\ \mathrm A \cos\theta-\mathrm C\sin\theta=0\vphantom{\dfrac{\mathrm A}{\sqrt{\mathrm A^{2}+\mathrm C^{2}}}} \end{cases} \right\} \tag{11}\label{eq11}$$

(3) Pauli matrices properties used : \begin{align} \sigma_{x}\sigma_{y} & = \boldsymbol{-}\sigma_{y}\sigma_{x}=i\sigma_{z}\,, \quad \sigma_{y} \sigma_{x}\sigma_{y} =\boldsymbol{-}\sigma_{x} \tag{12a}\label{eq12a}\\ \sigma_{y}\sigma_{z} & = \boldsymbol{-}\sigma_{z}\sigma_{y}=i\sigma_{x}\,, \quad \sigma_{y} \sigma_{z}\sigma_{y} =\boldsymbol{-}\sigma_{z} \tag{12b}\label{eq12b} \end{align}

• Thanks, @Frobenius for an illustrative explanation. Could you provide some reference for the bijection(one-to-one and onto correspondence) you mentioned? Jul 20, 2018 at 15:16
• @Akhil Pratap Singh : I don't have a specific reference. I learned about this self-studing about the representation of rotations by unitary matrices SU(2). Take a look for this subject in my answer here : Why is there this relationship between quaternions and Pauli matrices? Jul 20, 2018 at 15:32
• Why does the second term containing $\sigma_x$ vanishes in the last step of the Eq.(4)? Jul 20, 2018 at 16:29
• @Akhil Pratap Singh : $$\dfrac{\sin\theta}{\cos\theta}=\tan\theta \stackrel{defin. (03)}{=\!=\!=\!}\dfrac{\mathrm A }{\mathrm C }\Longrightarrow \mathrm A \cos\theta-\mathrm C\sin\theta=0$$ see also the footnote (2) at the end of answer. Jul 20, 2018 at 19:27