Question:
What is the wavefunction of the first excited state of the quantum harmonic oscillator, $|1\rangle$?
Solution:
The answer can easily be read from the Quantum Harmonic Oscillator page from Wikipedia. But since this is a blog of doing exercise, what is the point of copying the answer? Let us derive it from the ground state using the creation operator.
First, we have the ground state wavefunction $ \psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}} $.
Now we apply the creation operator $ \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}(\hat{x} - \frac{i\hat{p}}{m\omega}) = \sqrt{\frac{m\omega}{2\hbar}}(x - \frac{\hbar}{m\omega}\frac{\partial}{\partial x}) $.
Let's first compute the derivative:
$ \frac{\partial}{\partial x} \psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}}\frac{-m\omega x}{\hbar} $
So we have the momentum term:
$ - \frac{\hbar}{m\omega}\frac{\partial}{\partial x}\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}}x = x\psi_0(x) $
The position term is of course also $ x\psi_0(x) $. Combining using the creation operator we have
$ \hat{a}^\dagger \psi(x) = \sqrt{\frac{m\omega}{2\hbar}}(2x\psi_0(x)) = \frac{1}{\sqrt{2}}\psi_0(x)H_1(\frac{m\omega x}{\hbar}) $.
$ H_1 $ is the physicist Hermite polynomial of order 1 which is simply $ H_1(x) = 2x $.
What is the wavefunction of the first excited state of the quantum harmonic oscillator, $|1\rangle$?
Solution:
The answer can easily be read from the Quantum Harmonic Oscillator page from Wikipedia. But since this is a blog of doing exercise, what is the point of copying the answer? Let us derive it from the ground state using the creation operator.
First, we have the ground state wavefunction $ \psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}} $.
Now we apply the creation operator $ \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}(\hat{x} - \frac{i\hat{p}}{m\omega}) = \sqrt{\frac{m\omega}{2\hbar}}(x - \frac{\hbar}{m\omega}\frac{\partial}{\partial x}) $.
Let's first compute the derivative:
$ \frac{\partial}{\partial x} \psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}}\frac{-m\omega x}{\hbar} $
So we have the momentum term:
$ - \frac{\hbar}{m\omega}\frac{\partial}{\partial x}\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}}x = x\psi_0(x) $
The position term is of course also $ x\psi_0(x) $. Combining using the creation operator we have
$ \hat{a}^\dagger \psi(x) = \sqrt{\frac{m\omega}{2\hbar}}(2x\psi_0(x)) = \frac{1}{\sqrt{2}}\psi_0(x)H_1(\frac{m\omega x}{\hbar}) $.
$ H_1 $ is the physicist Hermite polynomial of order 1 which is simply $ H_1(x) = 2x $.
No comments:
Post a Comment