User Tools
notes:physics:quantum-mechanics:harmonic-oscillator
Differences
This shows you the differences between two versions of the page.
| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| notes:physics:quantum-mechanics:harmonic-oscillator [2026/09/27 20:47] – [The Number operator] chris | notes:physics:quantum-mechanics:harmonic-oscillator [2026/09/30 19:21] (current) – [position and momentum operators] chris | ||
|---|---|---|---|
| Line 437: | Line 437: | ||
| e^{-\large\frac{m\omega}{2\hbar}\normalsize x^2} | e^{-\large\frac{m\omega}{2\hbar}\normalsize x^2} | ||
| }$$ | }$$ | ||
| + | |||
| + | |||
| + | $$$$ | ||
| + | |||
| ==== The Number Operator ==== | ==== The Number Operator ==== | ||
| Line 494: | Line 498: | ||
| & | & | ||
| \end{align*}$$ | \end{align*}$$ | ||
| + | |||
| + | $$$$ | ||
| + | |||
| + | |||
| ==== Excited states ==== | ==== Excited states ==== | ||
| - | ==== How the construction | + | It was mentioned before that the creation and annihilation operators are called that because they |
| + | create or annihilate a quantum of energy $\hbar\omega$ respectively. So, when acting on an energy | ||
| + | eigenstate $\left|n\right> | ||
| + | |||
| + | Let's see how we can create excited states starting from the ground state. | ||
| + | |||
| + | $$\left|1\right> | ||
| + | |||
| + | We can find the energy level through the Number Operator. | ||
| + | |||
| + | $$ | ||
| + | \hat N\left|1\right> | ||
| + | = \hat N\hat a^\dagger\left|0\right> | ||
| + | =\left(\hat a^\dagger\hat N+\hat a^\dagger\right)\left|0\right> | ||
| + | =\hat a^\dagger\left|0\right> | ||
| + | \Leftrightarrow N=1 | ||
| + | $$ | ||
| + | $$E_1=\hbar\omega\left(N+\frac12\right)=\frac32\hbar\omega$$ | ||
| + | |||
| + | It just happens that it's already normalized. | ||
| + | |||
| + | $$\begin{align} | ||
| + | \left< | ||
| + | &= \left< | ||
| + | = \left< | ||
| + | =\left< | ||
| + | &= \big(\hat a\left|0\right> | ||
| + | \end{align}$$ | ||
| + | |||
| + | This doesn' | ||
| + | with $\left|k\right>: | ||
| + | |||
| + | $$\begin{align} | ||
| + | \left< k|k\right> | ||
| + | &= \left< | ||
| + | &= \left< | ||
| + | &= \left< | ||
| + | + \left< | ||
| + | &= 0+k\cdot\left< | ||
| + | &= k\cdot\left(\left< | ||
| + | | ||
| + | &= k\cdot\left(0+(k-1)\cdot\left< | ||
| + | &= k(k-1)\cdot\left< | ||
| + | &= \ ...\\ | ||
| + | &= k!\ \left< | ||
| + | &= k! | ||
| + | \end{align}$$ | ||
| + | |||
| + | For the normalized eigenstate $\left|n\right> | ||
| + | is $N=1/ | ||
| + | |||
| + | $$\boxed{\left|n\right> | ||
| + | |||
| + | The energy is: | ||
| + | |||
| + | $$ | ||
| + | \hat N\left|n\right> | ||
| + | = \hat N\left(\hat a^\dagger\right)^n\left|0\right> | ||
| + | = \left(\left(\hat a^\dagger\right)^n\hat N+n\left(\hat a^\dagger\right)^n\right)\left|0\right> | ||
| + | =n\left|n\right> | ||
| + | \Leftrightarrow N_n=n | ||
| + | $$ | ||
| + | |||
| + | $$\boxed{E_n=\hbar\omega\left(N_n+\frac12\right)=\left(n+\frac12\right)\hbar\omega}$$ | ||
| + | ==== Creation and Annihilation | ||
| + | |||
| + | While we can construct | ||
| + | be useful to go from one state to another one without having to pass through | ||
| + | the ground state. | ||
| + | |||
| + | If we notice that | ||
| + | $\require{cancel}\left[\hat a, | ||
| + | =\hat a\left(\hat a^\dagger\right)^n\left|0\right> | ||
| + | =\hat a\left(\hat a^\dagger\right)^n\left|0\right> | ||
| + | we can write: | ||
| + | |||
| + | $$\begin{align} | ||
| + | \hat a\left|n\right> | ||
| + | &= \frac{1}{\sqrt{n!}}\hat a\left(\hat a^\dagger\right)^n\left|0\right> | ||
| + | &= \frac{1}{\sqrt{n!}}\left[\hat a, | ||
| + | &= \frac{1}{\sqrt{n!}}n\left(\hat a^\dagger\right)^{n-1}\left|0\right> | ||
| + | &= \frac{1}{\sqrt{n!}}n\sqrt{(n-1)!}\left|n-1\right> | ||
| + | \\ | ||
| + | & | ||
| + | \end{align}$$ | ||
| + | |||
| + | Similarly, for the creation operator we have: | ||
| + | |||
| + | $$\begin{align} | ||
| + | \hat a^\dagger\left|n\right> | ||
| + | &= \frac{1}{\sqrt{n!}}\left(\hat a^\dagger\right)^{n+1}\left|0\right> | ||
| + | &= \frac{1}{\sqrt{n!}}\sqrt{(n+1)!}\left|n+1\right> | ||
| + | \\ | ||
| + | & | ||
| + | \end{align}$$ | ||
| + | ==== Position and Momentum operators ==== | ||
| + | |||
| + | The usefulness of the formulas above can be clearly seen while calculating the | ||
| + | expected values and the position and momentum dispersion. In a classical harmonic | ||
| + | oscillator the expected position should be the equilibrium point, $\left< x\right> | ||
| + | and the expected momentum should also be $\left< p\right> | ||
| + | $x$ and $p$ the oscillator is in, there is another //pair// state with $-x$ and $-p$. | ||
| + | The same happens in the quantum oscillator. | ||
| + | |||
| + | $$\begin{align} | ||
| + | \left< | ||
| + | &= \left< m\left|\hat x\right|n\right> | ||
| + | &= \sqrt{\frac{\hbar}{2m\omega}}\left< | ||
| + | &= \sqrt{\frac{\hbar}{2m\omega}}\left(\left< | ||
| + | &= \sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n+1}\left< | ||
| + | &= \sqrt{\frac{\hbar}{2m\omega}} (0+0)=0 | ||
| + | \end{align}$$ | ||
| + | |||
| + | $$\begin{align} | ||
| + | \left< | ||
| + | &= \left< n\left|\hat p\right|n\right> | ||
| + | &= i\sqrt{\frac{\hbar m\omega}{2}}\left< | ||
| + | &= i\sqrt{\frac{\hbar m\omega}{2}}\left(\left< | ||
| + | &= i\sqrt{\frac{\hbar m\omega}{2}}\left(\sqrt{n+1}\left< | ||
| + | &= i\sqrt{\frac{\hbar m\omega}{2}}(0+0)=0 | ||
| + | \end{align}$$ | ||
| + | |||
| + | Unlike the expected position and momentum, we cannot //guess// their dispersion. Here we | ||
| + | actually need the creation and annihilation operators in order to avoid difficult integrals. | ||
| + | |||
| + | $$\require{cancel}\begin{align} | ||
| + | \Delta x_n^2 | ||
| + | &= \left< | ||
| + | &= \left< n\left|\hat x\hat x\right|n\right> | ||
| + | &= \frac{\hbar}{2m\omega}\left< | ||
| + | &= \frac{\hbar}{2m\omega}\left< | ||
| + | \end{align}$$ | ||
| + | |||
| + | The operator $\hat a^\dagger\hat a^\dagger$ will give us the state $\left|n+2\right> | ||
| + | and it has no contribution to the sum. Similarly, we can ignore the $\hat a\hat a$ term. The operator $\hat a^\dagger\hat a$ is | ||
| + | just the number operator $\hat N$ with eigenvalue $n$. The operator $\hat a\hat a^\dagger$ can be written a different way | ||
| + | with the help of the commutator, $\hat a\hat a^\dagger=\left[\hat a,\hat a^\dagger\right]+\hat a^\dagger\hat a=1+\hat N$, and | ||
| + | then also replaced by the eigenvalue $1+n$. | ||
| + | |||
| + | $$\begin{align} | ||
| + | \Delta x_n^2 | ||
| + | &= \frac{\hbar}{2m\omega}\left< | ||
| + | &= \frac{\hbar}{2m\omega}\left(2n+1\right)\left< | ||
| + | &= \frac{\hbar}{2m\omega}\left(n+\frac12\right) | ||
| + | \end{align}$$ | ||
| + | |||
| + | When calculating $\Delta p_n^2$ we can use the same // | ||
| + | |||
| + | $$\require{cancel}\begin{align} | ||
| + | \Delta p_n^2 | ||
| + | &= \left< | ||
| + | &= \left< n\left|\hat p\hat p\right|n\right> | ||
| + | &= i^2\frac{\hbar m\omega}{2}\left< | ||
| + | &= \frac{\hbar m\omega}{2}\left< | ||
| + | &= \frac{\hbar m\omega}{2}\left< | ||
| + | &= \frac{\hbar m\omega}{2}\left< | ||
| + | &= \frac{\hbar m\omega}{2}\left(2n+1\right)\left< | ||
| + | &= \hbar m\omega\left(n+\frac12\right) | ||
| + | \end{align}$$ | ||
| - | ==== position and momentum operators ==== | + | We find the least dispersion at $n=0$. We then have: |
| + | $$\Delta x\Delta p=\sqrt{\frac12\frac{\hbar}{m\omega}}\cdot\sqrt{\frac12\hbar m\omega}=\frac{\hbar}{2}$$ | ||
| + | This result very much reminds us of [[uncertainty-principle|Heisenberg' | ||
notes/physics/quantum-mechanics/harmonic-oscillator.1790542029.txt.gz · Last modified: by chris
Except where otherwise noted, content on this wiki is licensed under the following license: CC Attribution-Share Alike 4.0 International
