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--- notes:physics:quantum-mechanics:harmonic-oscillator [2026/09/27 20:12] – [Natural Units] chris
+++ notes:physics:quantum-mechanics:harmonic-oscillator [2026/09/30 19:21] (current) – [position and momentum operators] chris
@@ Line 359: / Line 359: @@
 &\Leftrightarrow
 \left[\hat a, \hat a^\dagger\right] = 1
-\end{align*}
+\end{align*}$$
-$$
+$$$$
-These are what we call the creation and annihilation operators $\hat a^\dagger, \hat a$.
+These are what we call the **creation** and **annihilation** operators $\hat a^\dagger, \hat a$.
 They are called this way because when they act on the Hamiltonian, they create or annihilate
 quanta of energy, one at a time, and thus raise or lower the energy state of the oscillator.
@@ Line 383: / Line 383: @@
 $$\boxed{\hat x=\sqrt{\frac{\hbar}{2m\omega}}\left(\hat a^\dagger+\hat a\right)}$$
 $$\boxed{\hat p=i\sqrt{\frac{\hbar m\omega}{2}}\left(\hat a^\dagger-\hat a\right)}$$
-==== Rewriting the hamiltonian ====
+==== Rewriting the Hamiltonian ====
+We can now write the Hamiltonian like this:
+$$\begin{align*}
+\hat H
+&= \frac12m\omega^2V^\dagger V+\frac12\hbar\omega \\
+&= \frac12m\omega^2\frac{2\hbar}{m\omega}\hat a^\dagger\hat a+\frac12\hbar\omega \\
+&= \hbar\omega\hat a^\dagger\hat a+\frac12\hbar\omega
+\end{align*}$$
+$$\boxed{\hat H=\hbar\omega\left(\hat a^\dagger\hat a+\frac12\right)}$$
+$$$$
+Here we can see for the first time that a **fundamental energy state** $E_0$ exists.
+$$\begin{align*}
+\left<\hat H\right>
+&= \left<\psi\left|\hat H\right|\psi\right> \\
+&= \left<\psi\left|
+   \hbar\omega\left(
+     \hat a^\dagger\hat a+\frac12
+   \right)
+   \right|\psi\right> \\
+&= \hbar\omega\left(
+     \left<\psi\left|\hat a^\dagger\hat a\right|\psi\right>
+     +\frac12\left<\psi|\psi\right>
+   \right) \\
+&= \hbar\omega\left(
+     \left<\hat a\psi\big|\hat a\psi\right>+\frac12
+   \right) \ge \frac{\hbar\omega}{2} = E_0
+\end{align*}$$
+Requiring that the particle can take this fundamental energy gives us
+the form of the wavefunction. In order for $\left<\phi|\phi\right>=0$
+it must be $\phi=0$ and thus $\hat a\psi_0(x)=0$.
+$$\begin{align*}
+\hat a\psi_0(x) &= 0 \\
+\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i\hat p}{m\omega}\right)\psi_0(x) &= 0 \\
+x\psi_0(x)+\frac{i}{m\omega}\cdot(-i\hbar)\frac{d\psi_0}{dx} &= 0 \\
+x\psi_0(x)+\frac{\hbar}{m\omega}\frac{d\psi_0}{dx} &= 0 \\
+\frac{d\psi_0}{dx} &= -\frac{m\omega}{\hbar}x\psi_0(x) \\
+\\
+\psi_0(x) &= Ce^{-\large\frac{m\omega}{2\hbar}\normalsize x^2}
+\end{align*}$$
+After proper normalization, the fundamental wavefunction is:
+$$\boxed{
+\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}
+e^{-\large\frac{m\omega}{2\hbar}\normalsize x^2}
+}$$
+$$$$
+==== The Number Operator ====
+The operator $\hat a^\dagger\hat a$ has a unique name. It's called the **number operator**
+and is denoted by $\hat N$. The Hamiltonian is also written
+$\hat H=\hbar\omega\left(\hat N+1/2\right)$. It's evident that they have the same eigenstates.
+Its function is to //label the energy level of a wavefunction//, meaning $\hat N\left|n\right>=n$.
+It can easily be shown that $\hat N\left|0\right>=\hat a^\dagger\left(\hat a\left|0\right>\right)=\hat a^\dagger\cdot0=0$,
+and thus:
+$$\require{cancel}
+\hat H\left|0\right> = \hbar\omega\left(
+\cancelto{0}{\hat N\left|0\right>} + \frac12\left|0\right>
+\right) = \frac{\hbar\omega}{2}\left|0\right>
+$$
+$$E_0=\frac{\hbar\omega}{2}$$
+It's useful to study the behaviour of the number operator. This can be done by calculating
+its commutators.
+$$
+\big[\hat N,\hat a\big]
+=\big[\hat a^\dagger\hat a,\hat a\big]
+=\hat a^\dagger\big[\hat a,\hat a\big]+\big[\hat a^\dagger,\hat a\big]\hat a
+=0-\big[\hat a,\hat a^\dagger\big]\hat a
+=-\hat a
+$$
+$$
+\big[\hat N,\hat a^\dagger\big]
+=\big[\hat a^\dagger\hat a, \hat a^\dagger\big]
+=\hat a^\dagger\big[\hat a,\hat a^\dagger\big]+\big[\hat a^\dagger,\hat a^\dagger\big]\hat a
+=\hat a^\dagger\cdot 1+0
+=\hat a^\dagger
+$$
+$$\begin{align*}
+\big[\hat N,\hat a^k\big]
+ &=\hat a^\dagger\big[\hat a,\hat a^k\big]+\big[\hat a^\dagger,\hat a^k\big]\hat a
+  =0+\Big(\hat a^\dagger\hat a^k-\hat a^k\hat a^\dagger\Big)\hat a\\
+ &=\Big(\hat a^\dagger\hat a\hat a^{k-1}-\hat a^k\hat a^\dagger\Big)\hat a
+  =\Big((\hat a\hat a^\dagger-1)\hat a^{k-1}-\hat a^k\hat a^\dagger\Big)\hat a\\
+ &=\Big(\hat a\hat a^\dagger\hat a^{k-1}-\hat a^k\hat a^\dagger-\hat a^{k-1}\Big)\hat a
+  =\Big(\hat a(\hat a\hat a^\dagger-1)\hat a^{k-2}-\hat a^k\hat a^\dagger-\hat a^{k-1}\Big)\hat a\\
+ &=\Big(\hat a^2\hat a^\dagger\hat a^{k-2}-\hat a^k\hat a^\dagger-2\hat a^{k-1}\Big)\hat a
+  =\big(\hat a^k\hat a^\dagger-\hat a^k\hat a^\dagger-k\hat a^{k-1}\Big)\hat a\\
+ &=-k\hat a^k
+\end{align*}$$
+$$\begin{align*}
+\big[\hat N,(\hat a^\dagger)^k\big]
+ &=\hat a^\dagger\big[\hat a,(\hat a^\dagger)^k\big]+\big[\hat a^\dagger,(\hat a^\dagger)^k\big]\hat a
+  =\hat a^\dagger\Big(\hat a(\hat a^\dagger)^k-(\hat a^\dagger)^k\hat a\Big)+0\\
+ &=\hat a^\dagger\Big((\hat a^\dagger)^k\hat a-(\hat a^\dagger)^k\hat a+k(\hat a^\dagger)^{k-1}\Big)
+  =\hat a^\dagger k(\hat a^\dagger)^{k-1}\\
+ &=k(\hat a^\dagger)^k
+\end{align*}$$
+$$$$
-==== The Number operator ====
 ==== Excited states ====
-==== How the construction and annihilation operators act on energy eigenstates ====
+It was mentioned before that the creation and annihilation operators are called that because they
+create or annihilate a quantum of energy $\hbar\omega$ respectively. So, when acting on an energy
+eigenstate $\left|n\right>$ they give us the next or the previous eigenstate $\left|n+1\right>,\left|n-1\right>$.
+Let's see how we can create excited states starting from the ground state.
+$$\left|1\right> := \hat a^\dagger\left|0\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> =\left|1\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<1|1\right>
+&= \left<0\left|\left(\hat a^\dagger\right)^\dagger\hat a^\dagger\right|0\right>
+= \left<0\left|\hat a\hat a^\dagger\right|0\right>
+=\left<0\left|\left(\hat a^\dagger\hat a+1\right)\right|0\right> \\
+&= \big(\hat a\left|0\right>\big)^\dagger\hat a\left|0\right> + \left<0|1|0\right> = 0+\left<0|0\right>=1
+\end{align}$$
+This doesn't happen in general. Below, the non-normalized eigenstate will be denoted
+with $\left|k\right>:=\left(\hat a^\dagger\right)^k\left|0\right>$.
+$$\begin{align}
+\left< k|k\right>
+&= \left<0\left|\hat a^k\left(\hat a^\dagger\right)^k\right|0\right> \\
+&= \left<0\left|\hat a^{k-1}\left(\left(\hat a^\dagger\right)^k\hat a+k\left(\hat a^\dagger\right)^{k-1}\right)\right|0\right> \\
+&= \left<0\left<\hat a^{k-1}\left(\hat a^\dagger\right)^k\hat a\right|0\right>
+   + \left<0\left|k\hat a^{k-1}\left(\hat a^\dagger\right)^{k-1}\right|0\right> \\
+&= 0+k\cdot\left<0\left|\hat a^{k-2}\left(\left(\hat a^\dagger\right)^{k-1}\hat a+(k-1)\left(\hat a^\dagger\right)^{k-2}\right)\right|0\right> \\
+&= k\cdot\left(\left<0\left|\hat a^{k-2}\left(\hat a^\dagger\right)^{k-1}\hat a\right|0\right>
+   +\left<0\left|\hat a^{k-2}(k-1)\left(\hat a^\dagger\right)^{k-2}\right|0\right>\right) \\
+&= k\cdot\left(0+(k-1)\cdot\left<0\left|\hat a^{k-2}\left(\hat a^\dagger\right)^{k-2}\right|0\right>\right) \\
+&= k(k-1)\cdot\left<0\left|\hat a^{k-2}\left(\hat a^\dagger\right)^{k-2}\right|0\right> \\
+&= \ ...\\
+&= k!\ \left<0|0\right>\\
+&= k!
+\end{align}$$
+For the normalized eigenstate $\left|n\right>=N\left|k\right>$ is should be $N^2k!=1$, and thus the normalization constant
+is $N=1/\sqrt{n!}, \left|n\right>=1/\sqrt{n!}\ \left|k\right>$.
+$$\boxed{\left|n\right>=\frac{1}{\sqrt{n!}}\left(\hat a^\dagger\right)^n\left|0\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}$$
+==== Creation and Annihilation operators acting on Eigenstates ====
+While we can construct eigenstates starting from the ground state, it would
+be useful to go from one state to another one without having to pass through
+the ground state.
+If we notice that
+$\require{cancel}\left[\hat a,\left(\hat a^\dagger\right)^n\right]\left|0\right>
+=\hat a\left(\hat a^\dagger\right)^n\left|0\right> - \left(\hat a^\dagger\right)^n\cancelto{0}{\hat a\left|0\right>}
+=\hat a\left(\hat a^\dagger\right)^n\left|0\right>$
+we can write:
+$$\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,\left(\hat a^\dagger\right)^n\right]\left|0\right> \\
+&= \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>
+\\
+&\hspace{-21.5pt}\boxed{\hat a\left|n\right> = \sqrt{n}\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>
+\\
+&\hspace{-25pt}\boxed{\hat a^\dagger\left|n\right>=\sqrt{n+1}\left|n+1\right>}
+\end{align}$$
+==== Position and Momentum operators ====
+The usefulness of the formulas above can be clearly seen while calculating the
+expected values and the position and momentum dispersion. In a classical harmonic
+oscillator the expected position should be the equilibrium point, $\left< x\right>=0$,
+and the expected momentum should also be $\left< p\right>=0$, since for each state with
+$x$ and $p$ the oscillator is in, there is another //pair// state with $-x$ and $-p$.
+The same happens in the quantum oscillator.
+$$\begin{align}
+\left<\hat x\right>_n
+&= \left< m\left|\hat x\right|n\right> \\
+&= \sqrt{\frac{\hbar}{2m\omega}}\left< n\left|\hat a^\dagger+\hat a\right|n\right> \\
+&= \sqrt{\frac{\hbar}{2m\omega}}\left(\left< n\left|\hat a^\dagger\right|n\right> + \left< n\left|\hat a\right|n\right>\right) \\
+&= \sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n+1}\left< n\left|n+1\right.\right>+\sqrt{n}\left< n\left|n-1\right.\right>\right) \\
+&= \sqrt{\frac{\hbar}{2m\omega}} (0+0)=0
+\end{align}$$
+$$\begin{align}
+\left<\hat p\right>_n
+&= \left< n\left|\hat p\right|n\right> \\
+&= i\sqrt{\frac{\hbar m\omega}{2}}\left< n\left|\hat a^\dagger-\hat a\right|n\right> \\
+&= i\sqrt{\frac{\hbar m\omega}{2}}\left(\left< n\left|\hat a^\dagger\right|n\right>-\left< n\left|\hat a\right|n\right>\right) \\
+&= i\sqrt{\frac{\hbar m\omega}{2}}\left(\sqrt{n+1}\left< n\left|n+1\right.\right> - \sqrt{n}\left< n\left|n-1\right>\right>\right) \\
+&= 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<\hat x^2\right>_n-\cancelto{0}{\left<\hat x\right>_n^2} \\
+&= \left< n\left|\hat x\hat x\right|n\right> \\
+&= \frac{\hbar}{2m\omega}\left< n\left|\left(\hat a^\dagger+\hat a\right)\left(\hat a^\dagger+\hat a\right)\right|n\right> \\
+&= \frac{\hbar}{2m\omega}\left< n\left|\hat a^\dagger\hat a^\dagger+\hat a^\dagger\hat a+\hat a\hat a^\dagger+\hat a\hat a\right|n\right>
+\end{align}$$
+The operator $\hat a^\dagger\hat a^\dagger$ will give us the state $\left|n+2\right>$, which is perpendicular to $\left|n\right>$
+and it has no contribution to the sum. Similarly, we can ignore the $\hat a\hat a$ term. The operator $\hat a^\dagger\hat a$ is
+just the number operator $\hat N$ with eigenvalue $n$. The operator $\hat a\hat a^\dagger$ can be written a different way
+with the help of the commutator, $\hat a\hat a^\dagger=\left[\hat a,\hat a^\dagger\right]+\hat a^\dagger\hat a=1+\hat N$, and
+then also replaced by the eigenvalue $1+n$.
+$$\begin{align}
+\Delta x_n^2
+&= \frac{\hbar}{2m\omega}\left< n\left|2n+1\right|n\right> \\
+&= \frac{\hbar}{2m\omega}\left(2n+1\right)\left< n|n\right> \\
+&= \frac{\hbar}{2m\omega}\left(n+\frac12\right)
+\end{align}$$
+When calculating $\Delta p_n^2$ we can use the same //trick//:
+$$\require{cancel}\begin{align}
+\Delta p_n^2
+&= \left<\hat p^2\right>_n-\cancelto{0}{\left<\hat p\right>_n^2} \\
+&= \left< n\left|\hat p\hat p\right|n\right> \\
+&= i^2\frac{\hbar m\omega}{2}\left< n\left|\left(\hat a^\dagger-\hat a\right)\left(\hat a^\dagger-\hat a\right)\right|n\right> \\
+&= \frac{\hbar m\omega}{2}\left< n\left|\left(\hat a^\dagger-\hat a\right)\left(\hat a-\hat a^\dagger\right)\right|n\right> \\
+&= \frac{\hbar m\omega}{2}\left< n\left|\hat a^\dagger\hat a-\cancel{\hat a^\dagger\hat a^\dagger}-\cancel{\hat a\hat a}+\hat a\hat a^\dagger\right|n\right> \\
+&= \frac{\hbar m\omega}{2}\left< n\left|n+n+1\right|n\right> \\
+&= \frac{\hbar m\omega}{2}\left(2n+1\right)\left< n\left|n\right.\right> \\
+&= \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's uncertainty principle]].

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