Question:
Compute the root mean square of electron velocity in ground state quantized motion under the Earth's constant magnetic field.
Solution:
The problem seems daunting at first. But experience tell me once again this is most likely just number substituting exercise, so let's try expanding $ \hat{a}^\dagger \hat{a} $ and see what is going on there.
$ \begin{eqnarray*} & & \hat{a}^\dagger \hat{a} \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1} - i\hat{v_2})(\hat{v_1} + i\hat{v_2}) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i[\hat{v_1}, \hat{v_2}]) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i(i\frac{\hbar\omega}{m})) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) \\ \end{eqnarray*} $
So we can expand $ H_{2D} $ as follow:
$ \begin{eqnarray*} & & H_{2D} \\ &=& \hbar\omega(\hat{a}^\dagger \hat{a} + \frac{1}{2}) \\ &=& \hbar\omega(\frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{1}{2}) \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) - \frac{\hbar\omega}{2} + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) \\ \end{eqnarray*} $
Now we will equate the ground state energies
$ \begin{eqnarray*} \epsilon_0 | \psi_0 \rangle &=& H_{2D} | \psi_0 \rangle \\ \langle \psi_0 | \epsilon_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 \langle \psi_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \frac{m}{2}\langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{m} &=& \langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \end{eqnarray*} $
Last but not least, the actual value root mean square value is 32.15.
Compute the root mean square of electron velocity in ground state quantized motion under the Earth's constant magnetic field.
Solution:
The problem seems daunting at first. But experience tell me once again this is most likely just number substituting exercise, so let's try expanding $ \hat{a}^\dagger \hat{a} $ and see what is going on there.
$ \begin{eqnarray*} & & \hat{a}^\dagger \hat{a} \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1} - i\hat{v_2})(\hat{v_1} + i\hat{v_2}) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i[\hat{v_1}, \hat{v_2}]) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i(i\frac{\hbar\omega}{m})) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) \\ \end{eqnarray*} $
So we can expand $ H_{2D} $ as follow:
$ \begin{eqnarray*} & & H_{2D} \\ &=& \hbar\omega(\hat{a}^\dagger \hat{a} + \frac{1}{2}) \\ &=& \hbar\omega(\frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{1}{2}) \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) - \frac{\hbar\omega}{2} + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) \\ \end{eqnarray*} $
Now we will equate the ground state energies
$ \begin{eqnarray*} \epsilon_0 | \psi_0 \rangle &=& H_{2D} | \psi_0 \rangle \\ \langle \psi_0 | \epsilon_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 \langle \psi_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \frac{m}{2}\langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{m} &=& \langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \end{eqnarray*} $
Last but not least, the actual value root mean square value is 32.15.
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