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--- notes:physics:quantum-mechanics:harmonic-oscillator [2026/09/27 16:36] – [Trying to guess the operators] chris
+++ notes:physics:quantum-mechanics:harmonic-oscillator [2026/09/30 19:21] (current) – [position and momentum operators] chris
@@ Line 43: / Line 43: @@
 $$\psi''(x)+\left(2E-x^2\right)\psi(x)=0 \tag{2}$$
-The *problem* with this approach when first learning about the quantum harmonic
+The //problem// with this approach when first learning about the quantum harmonic
 oscillator is that you kind have to already know before what type of result you
 expect and what unit reconstruction to do, or blindly trust your textbook, ending
-up in not understanding what you're doing (*that's what happened to me at least*).
+up in not understanding what you're doing (//that's what happened to me at least//).
 The second approach is starting by demanding the use of a dimensionless variable, and
@@ Line 345: / Line 345: @@
 ==== Construction & Annihilation operators ====
-==== Rewriting the hamiltonian ====
+Ideally, we'd like to have these operators, $V^\dagger$ and $V$, in some type of normalized
+form. We can achieve that by studying their commutator.
+$$\begin{align*}
+\left[V,V^\dagger\right]
+&= \left[\hat x+\frac{i\hat p}{m\omega},\hat x-\frac{i\hat p}{m\omega}\right] \\
+&= -\frac{i}{m\omega}\left[\hat x,\hat p\right] + \frac{i}{m\omega}\left[\hat p,\hat x\right] \\
+&= -\frac{i}{m\omega}\big(i\hbar-(-i\hbar)\big)\\
+&= \frac{2\hbar}{m\omega} \\
+\\
+&\Leftrightarrow
+\left[\sqrt{\frac{m\omega}{2\hbar}}V,\sqrt{\frac{m\omega}{2\hbar}}V^\dagger\right]=1 \\
+&\Leftrightarrow
+\left[\hat a, \hat a^\dagger\right] = 1
+\end{align*}$$
+$$$$
+These are what we call the **creation** and **annihilation** operators $\hat a^\dagger, \hat a$.
+They are called this way because when they act on the Hamiltonian, they create or annihilate
+quanta of energy, one at a time, and thus raise or lower the energy state of the oscillator.
+$$\boxed{\hat a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x-\frac{i}{m\omega}\hat p\right)}$$
+$$\boxed{\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i}{m\omega}\hat p\right)}$$
+$$\left[\hat a,\hat a^\dagger\right]=1$$
+$$\left[\hat a,\left(\hat a^\dagger\right)^n\right] = n\left(\hat a^\dagger\right)^{n-1}$$
+Obviously those operators are not hermitian. But their combination is, just like $V^\dagger$ and $V$,
+since $\left(\hat a^\dagger\hat a\right)=\hat a^\dagger\left(\hat a^\dagger\right)^\dagger = \hat a^\dagger\hat a$.
+This factorization was first used by Paul Dirac and isn't limited to just the quantum harmonic oscillator.
+But their names are specific to the QHO, due to the quanta creation/annihilation. When used more generally
+they are called **ladder operators** as a pair, or **raising** and **lowering** operators when referring to
+them individually.
+It is also useful to solve the system above for the position and momentum operators.
+$$\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)}$$
+==== Rewriting the Hamiltonian ====
+We can now write the Hamiltonian like this:
+$$\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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