$ \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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Significance

Derivation

We will look at an example for an electron. In order to look at the time evolution in a quantum system, we need the Hamiltonian. Recall for spin, we have that $\vec{\mu}=g\frac{q}{2m}\vec{s}$.
The Hamiltonian represents total energy. Here the energy is only provided by the magnetic field and hence we have $$\begin{align*} H&=-\vec{\mu}\cdot\vec{B} \\ &=-g\frac{q}{2m}\vec{s}\cdot\vec{B} \\ &=\frac{2}{m_e}\vec{s}\cdot\vec{B}. \end{align*}$$ We consider the magnetic field along $x$, which means we can write $\vec{B}=B\vec{z}$. Then $$\begin{align*} H=\frac{eB}{m_e}S_z. \end{align*}$$
Let's gather the scalar multiple into $\omega_0=\frac{eB}{m_e}$, which are called Larmour frequency. Since $|+\rangle$, $|-\langle$ are eigenstates of $S_z$, and $H$ is a scalar multiple of $S_z$, then they are also energy eigenstates of $H$ and so the eigenvalue equations of $H$ is $$\begin{align*} H|+\rangle = omega_0S_z|+\rangle &= \frac{\hbar}{2}\omega|+\rangle \\ H|-\rangle = \omega_0S_z|-\rangle &=-\frac{\hbar}{2}\omega_0|-\rangle, \end{align*}$$ where $\pm\frac{\hbar}{2}\omega$ are the eigenvalues.
Let's initialize the system in a general state $$\begin{align*} |+\rangle_n=\cos(\frac{\theta}{2})|+\rangle+ e^{i\phi}\sin(\frac{\theta}{2})|-\rangle \end{align*}$$ that is, $|\psi(0)\rangle=|+\rangle_n$, where $|+\rangle_n$ can be represented in matrix form as $$\begin{align*} |+\rangle_n=\begin{bmatrix}\cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix} \end{align*}$$
We can compare this to a classical case
Recall a classic magnetic moment, where $\mu=\frac{q}{2m}\vec{L}$. A torque comes from a magneti field of where $\vec{\tau}=\vec{\mu}\times\vec{B}$, where $\vec{B}$ is the force of the magnetic field.
Since $\frac{d\vec{L}}{dt}=\vec{\tau}=\vec{mu}\times\vec{B}$, we can muliply both sides by $\frac{q}{2m}$ to get $$\begin{align*} \frac{d\vec{u}}{dt}=

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Significance

Derivation

We will look at an example for an electron. In order to look at the time evolution in a quantum system, we need the Hamiltonian. Recall for spin, we have that $\vec{\mu}=g\frac{q}{2m}\vec{s}$.
The Hamiltonian represents total energy. Here the energy is only provided by the magnetic field and hence we have $$\begin{align*} H&=-\vec{\mu}\cdot\vec{B} \\ &=-g\frac{q}{2m}\vec{s}\cdot\vec{B} \\ &=\frac{2}{m_e}\vec{s}\cdot\vec{B}. \end{align*}$$ We consider the magnetic field along $x$, which means we can write $\vec{B}=B\vec{z}$. Then $$\begin{align*} H=\frac{eB}{m_e}S_z. \end{align*}$$
Let's gather the scalar multiple into $\omega_0=\frac{eB}{m_e}$, which are called Larmour frequency. Since $|+\rangle$, $|-\langle$ are eigenstates of $S_z$, and $H$ is a scalar multiple of $S_z$, then they are also energy eigenstates of $H$ and so the eigenvalue equations of $H$ is $$\begin{align*} H|+\rangle = omega_0S_z|+\rangle &= \frac{\hbar}{2}\omega|+\rangle \\ H|-\rangle = \omega_0S_z|-\rangle &=-\frac{\hbar}{2}\omega_0|-\rangle, \end{align*}$$ where $\pm\frac{\hbar}{2}\omega$ are the eigenvalues.
Let's initialize the system in a general state $$\begin{align*} |+\rangle_n=\cos(\frac{\theta}{2})|+\rangle+ e^{i\phi}\sin(\frac{\theta}{2})|-\rangle \end{align*}$$ that is, $|\psi(0)\rangle=|+\rangle_n$, where $|+\rangle_n$ can be represented in matrix form as $$\begin{align*} |+\rangle_n=\begin{bmatrix}\cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix} \end{align*}$$
We can compare this to a classical case
Recall a classic magnetic moment, where $\mu=\frac{q}{2m}\vec{L}$. A torque comes from a magneti field of where $\vec{\tau}=\vec{\mu}\times\vec{B}$, where $\vec{B}$ is the force of the magnetic field.
Since $\frac{d\vec{L}}{dt}=\vec{\tau}=\vec{mu}\times\vec{B}$, we can muliply both sides by $\frac{q}{2m}$ to get $$\begin{align*} \frac{d\vec{u}}{dt}=

Related Experiments

Coming soon

Associated Concepts

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

Significance

Derivation

We will look at an example for an electron. In order to look at the time evolution in a quantum system, we need the Hamiltonian. Recall for spin, we have that $\vec{\mu}=g\frac{q}{2m}\vec{s}$.
The Hamiltonian represents total energy. Here the energy is only provided by the magnetic field and hence we have $$\begin{align*} H&=-\vec{\mu}\cdot\vec{B} \\ &=-g\frac{q}{2m}\vec{s}\cdot\vec{B} \\ &=\frac{2}{m_e}\vec{s}\cdot\vec{B}. \end{align*}$$ We consider the magnetic field along $x$, which means we can write $\vec{B}=B\vec{z}$. Then $$\begin{align*} H=\frac{eB}{m_e}S_z. \end{align*}$$
Let's gather the scalar multiple into $\omega_0=\frac{eB}{m_e}$, which are called Larmour frequency. Since $|+\rangle$, $|-\langle$ are eigenstates of $S_z$, and $H$ is a scalar multiple of $S_z$, then they are also energy eigenstates of $H$ and so the eigenvalue equations of $H$ is $$\begin{align*} H|+\rangle = omega_0S_z|+\rangle &= \frac{\hbar}{2}\omega|+\rangle \\ H|-\rangle = \omega_0S_z|-\rangle &=-\frac{\hbar}{2}\omega_0|-\rangle, \end{align*}$$ where $\pm\frac{\hbar}{2}\omega$ are the eigenvalues.
Let's initialize the system in a general state $$\begin{align*} |+\rangle_n=\cos(\frac{\theta}{2})|+\rangle+ e^{i\phi}\sin(\frac{\theta}{2})|-\rangle \end{align*}$$ that is, $|\psi(0)\rangle=|+\rangle_n$, where $|+\rangle_n$ can be represented in matrix form as $$\begin{align*} |+\rangle_n=\begin{bmatrix}\cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix} \end{align*}$$
We can compare this to a classical case
Recall a classic magnetic moment, where $\mu=\frac{q}{2m}\vec{L}$. A torque comes from a magneti field of where $\vec{\tau}=\vec{\mu}\times\vec{B}$, where $\vec{B}$ is the force of the magnetic field.
Since $\frac{d\vec{L}}{dt}=\vec{\tau}=\vec{mu}\times\vec{B}$, we can muliply both sides by $\frac{q}{2m}$ to get $$\begin{align*} \frac{d\vec{u}}{dt}=

Related Experiments

Coming soon

Associated Concepts

Coming soon

Significance

Derivation

We will look at an example for an electron. In order to look at the time evolution in a quantum system, we need the Hamiltonian. Recall for spin, we have that $\vec{\mu}=g\frac{q}{2m}\vec{s}$.
The Hamiltonian represents total energy. Here the energy is only provided by the magnetic field and hence we have $$\begin{align*} H&=-\vec{\mu}\cdot\vec{B} \\ &=-g\frac{q}{2m}\vec{s}\cdot\vec{B} \\ &=\frac{2}{m_e}\vec{s}\cdot\vec{B}. \end{align*}$$ We consider the magnetic field along $x$, which means we can write $\vec{B}=B\vec{z}$. Then $$\begin{align*} H=\frac{eB}{m_e}S_z. \end{align*}$$
Let's gather the scalar multiple into $\omega_0=\frac{eB}{m_e}$, which are called Larmour frequency. Since $|+\rangle$, $|-\langle$ are eigenstates of $S_z$, and $H$ is a scalar multiple of $S_z$, then they are also energy eigenstates of $H$ and so the eigenvalue equations of $H$ is $$\begin{align*} H|+\rangle = omega_0S_z|+\rangle &= \frac{\hbar}{2}\omega|+\rangle \\ H|-\rangle = \omega_0S_z|-\rangle &=-\frac{\hbar}{2}\omega_0|-\rangle, \end{align*}$$ where $\pm\frac{\hbar}{2}\omega$ are the eigenvalues.
Let's initialize the system in a general state $$\begin{align*} |+\rangle_n=\cos(\frac{\theta}{2})|+\rangle+ e^{i\phi}\sin(\frac{\theta}{2})|-\rangle \end{align*}$$ that is, $|\psi(0)\rangle=|+\rangle_n$, where $|+\rangle_n$ can be represented in matrix form as $$\begin{align*} |+\rangle_n=\begin{bmatrix}\cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix} \end{align*}$$
We can compare this to a classical case
Recall a classic magnetic moment, where $\mu=\frac{q}{2m}\vec{L}$. A torque comes from a magneti field of where $\vec{\tau}=\vec{\mu}\times\vec{B}$, where $\vec{B}$ is the force of the magnetic field.
Since $\frac{d\vec{L}}{dt}=\vec{\tau}=\vec{mu}\times\vec{B}$, we can muliply both sides by $\frac{q}{2m}$ to get $$\begin{align*} \frac{d\vec{u}}{dt}=

Related Experiments

Coming soon

Associated Concepts

Coming soon
FullPage
overview
significance
derivation
related experiments
associated concepts