$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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overview
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Significance

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Derivation

We consider a magnetic field along two axis $\vec{B}=B_0\hat{z}+B_1\hat{x}$. Define two larmour frequences of $\omega_0=\frac{eB_0}{m_e}$ and $\omega_1=\frac{eB_1}{m_e}$.
The total energy is then $$\begin{align*} H&=-\muB_0\vec{z}-\muB_1\vec{x} \\ &=-g\frac{q}{2m}\vec{s_z}B_0\vec{z}-g\frac{q}{2m}\vec{s_x} \\ &=\frac{eB_0}{m_e}\vec{s_z}B_0\vec{z} +\frac{eB_1}{m_e}\vec{s_x}B_1\vec{x} \\ &=\omega_0S_z+\omega_1S_x \\ &=\frac{\hbar}{2}\begin{bmatrix}\omega_0 & \omega_1 \\ \omega_1 & -\omega_0\end{bmatrix} \end{align*}$$ To find the possible results and resultant states of this measurement, we need the eigenvalues and eigenstates of the Hamiltonian.

Associated Concepts

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Significance

Coming soon

Derivation

We consider a magnetic field along two axis $\vec{B}=B_0\hat{z}+B_1\hat{x}$. Define two larmour frequences of $\omega_0=\frac{eB_0}{m_e}$ and $\omega_1=\frac{eB_1}{m_e}$.
The total energy is then $$\begin{align*} H&=-\muB_0\vec{z}-\muB_1\vec{x} \\ &=-g\frac{q}{2m}\vec{s_z}B_0\vec{z}-g\frac{q}{2m}\vec{s_x} \\ &=\frac{eB_0}{m_e}\vec{s_z}B_0\vec{z} +\frac{eB_1}{m_e}\vec{s_x}B_1\vec{x} \\ &=\omega_0S_z+\omega_1S_x \\ &=\frac{\hbar}{2}\begin{bmatrix}\omega_0 & \omega_1 \\ \omega_1 & -\omega_0\end{bmatrix} \end{align*}$$ To find the possible results and resultant states of this measurement, we need the eigenvalues and eigenstates of the Hamiltonian.

Associated Concepts

Coming soon
FullPage
overview
significance
derivation
related experiments
associated concepts
FullPage
overview
significance
derivation
related experiments
associated concepts

Significance

Coming soon

Derivation

We consider a magnetic field along two axis $\vec{B}=B_0\hat{z}+B_1\hat{x}$. Define two larmour frequences of $\omega_0=\frac{eB_0}{m_e}$ and $\omega_1=\frac{eB_1}{m_e}$.
The total energy is then $$\begin{align*} H&=-\muB_0\vec{z}-\muB_1\vec{x} \\ &=-g\frac{q}{2m}\vec{s_z}B_0\vec{z}-g\frac{q}{2m}\vec{s_x} \\ &=\frac{eB_0}{m_e}\vec{s_z}B_0\vec{z} +\frac{eB_1}{m_e}\vec{s_x}B_1\vec{x} \\ &=\omega_0S_z+\omega_1S_x \\ &=\frac{\hbar}{2}\begin{bmatrix}\omega_0 & \omega_1 \\ \omega_1 & -\omega_0\end{bmatrix} \end{align*}$$ To find the possible results and resultant states of this measurement, we need the eigenvalues and eigenstates of the Hamiltonian.

Associated Concepts

Coming soon

Significance

Coming soon

Derivation

We consider a magnetic field along two axis $\vec{B}=B_0\hat{z}+B_1\hat{x}$. Define two larmour frequences of $\omega_0=\frac{eB_0}{m_e}$ and $\omega_1=\frac{eB_1}{m_e}$.
The total energy is then $$\begin{align*} H&=-\muB_0\vec{z}-\muB_1\vec{x} \\ &=-g\frac{q}{2m}\vec{s_z}B_0\vec{z}-g\frac{q}{2m}\vec{s_x} \\ &=\frac{eB_0}{m_e}\vec{s_z}B_0\vec{z} +\frac{eB_1}{m_e}\vec{s_x}B_1\vec{x} \\ &=\omega_0S_z+\omega_1S_x \\ &=\frac{\hbar}{2}\begin{bmatrix}\omega_0 & \omega_1 \\ \omega_1 & -\omega_0\end{bmatrix} \end{align*}$$ To find the possible results and resultant states of this measurement, we need the eigenvalues and eigenstates of the Hamiltonian.

Associated Concepts

Coming soon
FullPage
overview
significance
derivation
related experiments
associated concepts