The Schrödinger Equation and Describing the Proton's State

• Proton is going to be in either the up state or down state, and is going to transition back and forth. Need the math to understand this. Describe state with function $\Psi (\vec{r},t)$. Set up function to have certain properties
  

  $$\int_{\Delta V} \vert\Psi (\vec{r},t)\vert^{2} dV = {\rm pro... ...\;the\;particle\; in \; volume \; \Delta V \; at \; time \; t}$$ (5.6)

  
  

  
  
  
  
  
  
  
  
  
  
  
  
  
  

• Schrödinger Equation describes how wave functions behave
  

  $$H \Psi = i \hbar \frac{\partial \Psi}{\partial t}$$

  $$\left(\frac{-\hbar ^{2}}{2m} \nabla ^{2} + U \right)\Psi = \frac{\partial \Psi}{\partial t}$$ (5.7)

  
  

• If $H$ is independent of spatial location, can re-write (5.7)
  

  $$U \Psi = \frac{\partial \Psi}{\partial t}$$ (5.8)

  
  

• Since $H$ is independent of time, can use separation of variables to express $\Psi$ as product
  

  $$\Psi (T) = \Psi e^{-\frac{i}{\hbar} E t}$$ (5.9)

  
  

• Plug (5.9) and (5.3) into (5.8) to get
  

  $$(-\mu \cdot \vec{B}) \Psi = E \Psi$$ (5.10)

  
  

• where we know $E = m_{s} \hbar \omega_{0}$, there are only two possible states to construct $\Psi$ from. Choose to use the Pauli representation of this problem. Use vectors to represent states and matrices to represent Hamiltonian.

• Two possible states for proton

  
  

  $$\psi_{+} \equiv \psi_{+1/2} = \left( \begin{array}{c} 1 \\ 0 \\ \end{array}\right) \hspace{1.0in} \mbox{ spin parallel (\lq spin up')}$$ (5.11)

  $$\psi_{-} \equiv \psi_{-1/2} = \left( \begin{array}{c} 0 \\ 1 \\ \end{array}\right) \hspace{1.0in} \mbox{ spin anti-parallel (\lq spin down')}$$ (5.12)

  
  

  and
  

  $$H = -\gamma B_0 S_z = \left( \begin{array}{cc} -\frac{1}{2} ... ...0 \\ 0 & +\frac{1}{2} \hbar \omega_{0} \\ \end{array}\right)$$ (5.13)

  
  

• Write it out for completeness

  
  

  $$\left( \begin{array}{cc} -\frac{1}{2} \hbar \omega_{0} & 0 \... ...end{array}\right) \Psi = \mp \frac{1}{2} \hbar \omega_{0} \Psi$$ (5.14)

  
  

• Wave function has to describe all possible states for the proton, so it will be a sum of the states
  

  $$\Psi(t) = \sum_{m = \pm\frac {1}{2} } C_{m}\psi_{m} e^{-\frac{i}{\hbar} E_{m}t} \\$$ (5.15)

  
  

• Proton has to exist implies normalization
  

  $$<\Psi\vert\Psi> \equiv \int \Psi^\dagger \Psi dV = \Psi^\da... ...E_{m})t} = 1 \;\;\;\;\;\;\;where\; E_{m} = -m\hbar \omega_{0}$$ (5.16)

  
  
  so
  

  $$V\sum_{m}\vert C_{m}\vert^{2} = 1 \\$$ (5.17)

  
  

• Now know how to describe proton, how do we get to precession.