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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 $$\begin{align*} \frac{d\vec{u}}{dt}=
Associated Concepts
Coming soonSignificance
Derivation
We will look at an example for an electron. In order to look at the time evolution in a quantum system, we need the Hamiltonian. Recall for spin, we have that $\vec{\mu}=g\frac{q}{2m}\vec{s}$.The Hamiltonian represents total energy. Here the energy is only provided by the magnetic field and hence we have $$\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}=
Associated Concepts
Coming soon
FullPage
overview
significance
derivation
associated concepts