Schrödinger equation \(\mathring{\imath}\)\(\hslash\)\(\left(\vphantom{\partial \,/\, \partial t}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(t\)\(\left.\vphantom{\partial \,/\, \partial t}\right)\)\(\psi\)\(\left[\vphantom{x ,\, t}\right.\)\(x\)\(,\,\)\(t\)\(\left.\vphantom{x ,\, t}\right]\)\(\;=\;\)\(H\)\(\psi\)\(\left[\vphantom{x ,\, t}\right.\)\(x\)\(,\,\)\(t\)\(\left.\vphantom{x ,\, t}\right]\) states a relation between wave function \(\psi\) and Hamiltonian \(H\), but it is unclear without empirical evidence what the expression of \(H\) should be.
Define the commutator:
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{A ,\, B}\right.\)\(A\)\(,\,\)\(B\)\(\left.\vphantom{A ,\, B}\right]\)\(\;=\;\)\(A\)\(B\)\(\,-\,\)\(B\)\(A\)\(;\,\)
Suppose observables \(A\)\(,\,\)\(B\) have \(\,\mathrm{Comm}\,\)\(\left[\vphantom{A ,\, B}\right.\)\(A\)\(,\,\)\(B\)\(\left.\vphantom{A ,\, B}\right]\)\(\;=\;\)\(0\), then they can be simultaneously diagonalized. A physical argument goes as thus. Let \(A\) be already diagonal with a suitable basis, then \(A \vphantom{ A}_{\left[m ,\, m\right]}\)\(B \vphantom{ B}_{\left[m ,\, n\right]}\)\(\;=\;\)\(\left(A B\right) \vphantom{ \left(A B\right)}_{\left[m ,\, n\right]}\)\(\;=\;\)\(\left(B A\right) \vphantom{ \left(B A\right)}_{\left[m ,\, n\right]}\)\(\;=\;\)\(A \vphantom{ A}_{\left[n ,\, n\right]}\)\(B \vphantom{ B}_{\left[m ,\, n\right]}\), in which case \(B\) is diagonal unless \(A\)\(\;=\;\)\(I\).
Quantum mechanics can be based on commutation relations, as Born did. For the operators of position \(X\)\(,\,\)\(Y\)\(,\,\)\(Z\) and momentum \(P \vphantom{ P}_{x}\)\(,\,\)\(P \vphantom{ P}_{y}\)\(,\,\)\(P \vphantom{ P}_{z}\) it says that:
Commutation_position:
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{X ,\, P \vphantom{ P}_{x}}\right.\)\(X\)\(,\,\)\(P \vphantom{ P}_{x}\)\(\left.\vphantom{X ,\, P \vphantom{ P}_{x}}\right]\)\(\;=\;\)\(\mathring{\imath}\)\(\hslash\)\(;\,\)
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{Y ,\, P \vphantom{ P}_{y}}\right.\)\(Y\)\(,\,\)\(P \vphantom{ P}_{y}\)\(\left.\vphantom{Y ,\, P \vphantom{ P}_{y}}\right]\)\(\;=\;\)\(\mathring{\imath}\)\(\hslash\)\(;\,\)
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{Z ,\, P \vphantom{ P}_{z}}\right.\)\(Z\)\(,\,\)\(P \vphantom{ P}_{z}\)\(\left.\vphantom{Z ,\, P \vphantom{ P}_{z}}\right]\)\(\;=\;\)\(\mathring{\imath}\)\(\hslash\)\(;\,\)
There are a variety of motivations for these relations. I suggest that we simply assume CommPos as an axiom. (Some authors postulate that the momentum and angular momentum operators are a generator of translation and rotation, and the commutation relations between angular momenta come from the fact that infinitesimal rotations are not commutative; but how do we continue the analogy to explain the time evolution of wave functions?) This is how I see it: In classical mechanics it doesn’t matter that you see the position or the momentum first, but in quantum mechanics, it not only matters, it modifies the target too, and such action is not commutative; this is to me the single most important message to remember. Now if \(\,\mathrm{Comm}\,\)\(\left[\vphantom{X ,\, P}\right.\)\(X\)\(,\,\)\(P\)\(\left.\vphantom{X ,\, P}\right]\)\(\;=\;\)\(C\) is diagonal and nowhere vanishes, we scale \(X\)\(,\,\)\(P\) such that \(\,\mathrm{Comm}\,\)\(\left[\vphantom{X ,\, P}\right.\)\(X\)\(,\,\)\(P\)\(\left.\vphantom{X ,\, P}\right]\)\(\;=\;\)\(c\)\(I\) by redefining physical units. Furthermore, \(c\) must be purely imaginary in order to be hermitian, which we take to be \(\mathring{\imath}\)\(\hslash\) .
In the positional representation, \(X\) is simply a multiplication of \(x\), and the expressions of \(X\)\(,\,\)\(P\) turns out to be:
Representation_momentum:
\(\mathbf{P}\)\(\;=\;\)\(\,-\,\)\(\mathring{\imath}\)\(\hslash\)\(\nabla\)\(;\,\)
This is easy to verify, but does CommPos determine RepMom uniquely? Yes, because if \(X\)\(P \vphantom{ P}_{1}\)\(\,-\,\)\(P \vphantom{ P}_{1}\)\(X\)\(\;=\;\)\(X\)\(P \vphantom{ P}_{2}\)\(\,-\,\)\(P \vphantom{ P}_{2}\)\(X\)\(\;=\;\)\(\mathring{\imath}\)\(\hslash\), then \(X\)\(\left(\vphantom{P \vphantom{ P}_{1} \,-\, P \vphantom{ P}_{2}}\right.\)\(P \vphantom{ P}_{1}\)\(\,-\,\)\(P \vphantom{ P}_{2}\)\(\left.\vphantom{P \vphantom{ P}_{1} \,-\, P \vphantom{ P}_{2}}\right)\)\(\,-\,\)\(\left(\vphantom{P \vphantom{ P}_{1} \,-\, P \vphantom{ P}_{2}}\right.\)\(P \vphantom{ P}_{1}\)\(\,-\,\)\(P \vphantom{ P}_{2}\)\(\left.\vphantom{P \vphantom{ P}_{1} \,-\, P \vphantom{ P}_{2}}\right)\)\(X\)\(\;=\;\)\(0\), and \(P \vphantom{ P}_{1}\)\(\,-\,\)\(P \vphantom{ P}_{2}\) must be a constant function.
In the three dimensional space, let \(\mathbf{X}\)\(,\,\)\(\mathbf{P}\) be the vector-valued position and momentum operators. There are additional commutation relations for orbital angular momenta \(L \vphantom{ L}_{x}\)\(,\,\)\(L \vphantom{ L}_{y}\)\(,\,\)\(L \vphantom{ L}_{z}\) (another two permuted relations also holds):
Commutation_angular_orbital:
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{L \vphantom{ L}_{x} ,\, L \vphantom{ L}_{y}}\right.\)\(L \vphantom{ L}_{x}\)\(,\,\)\(L \vphantom{ L}_{y}\)\(\left.\vphantom{L \vphantom{ L}_{x} ,\, L \vphantom{ L}_{y}}\right]\)\(\;=\;\)\(\mathring{\imath}\)\(\hslash\)\(L \vphantom{ L}_{z}\)\(;\,\)
If we guess \(L \vphantom{ L}_{z}\)\(\;=\;\)\(\,-\,\)\(\mathring{\imath}\)\(\hslash\)\(\left(\vphantom{\partial \,/\, \partial \varphi}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(\varphi\)\(\left.\vphantom{\partial \,/\, \partial \varphi}\right)\), with \(\varphi\) the azimuth angle, it can be recovered that:
Representation_angular_orbital:
\(\mathbf{L}\)\(\;=\;\)\(\mathbf{R}\)\(\,\times\,\)\(\mathbf{P}\)\(;\,\)
We can show CommAngOrb by directly using RepAngOrb and RepMom. To see this, verify that \(\left(\vphantom{y \left(\partial \,/\, \partial z\right) \,-\, z \left(\partial \,/\, \partial y\right)}\right.\)\(y\)\(\left(\vphantom{\partial \,/\, \partial z}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(z\)\(\left.\vphantom{\partial \,/\, \partial z}\right)\)\(\,-\,\)\(z\)\(\left(\vphantom{\partial \,/\, \partial y}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(y\)\(\left.\vphantom{\partial \,/\, \partial y}\right)\)\(\left.\vphantom{y \left(\partial \,/\, \partial z\right) \,-\, z \left(\partial \,/\, \partial y\right)}\right)\)\(\left(\vphantom{z \left(\partial \,/\, \partial x\right) \,-\, x \left(\partial \,/\, \partial z\right)}\right.\)\(z\)\(\left(\vphantom{\partial \,/\, \partial x}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(x\)\(\left.\vphantom{\partial \,/\, \partial x}\right)\)\(\,-\,\)\(x\)\(\left(\vphantom{\partial \,/\, \partial z}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(z\)\(\left.\vphantom{\partial \,/\, \partial z}\right)\)\(\left.\vphantom{z \left(\partial \,/\, \partial x\right) \,-\, x \left(\partial \,/\, \partial z\right)}\right)\)\(\,-\,\)\(\left(\vphantom{z \left(\partial \,/\, \partial x\right) \,-\, x \left(\partial \,/\, \partial z\right)}\right.\)\(z\)\(\left(\vphantom{\partial \,/\, \partial x}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(x\)\(\left.\vphantom{\partial \,/\, \partial x}\right)\)\(\,-\,\)\(x\)\(\left(\vphantom{\partial \,/\, \partial z}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(z\)\(\left.\vphantom{\partial \,/\, \partial z}\right)\)\(\left.\vphantom{z \left(\partial \,/\, \partial x\right) \,-\, x \left(\partial \,/\, \partial z\right)}\right)\)\(\left(\vphantom{y \left(\partial \,/\, \partial z\right) \,-\, z \left(\partial \,/\, \partial y\right)}\right.\)\(y\)\(\left(\vphantom{\partial \,/\, \partial z}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(z\)\(\left.\vphantom{\partial \,/\, \partial z}\right)\)\(\,-\,\)\(z\)\(\left(\vphantom{\partial \,/\, \partial y}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(y\)\(\left.\vphantom{\partial \,/\, \partial y}\right)\)\(\left.\vphantom{y \left(\partial \,/\, \partial z\right) \,-\, z \left(\partial \,/\, \partial y\right)}\right)\)\(\;=\;\)\(x\)\(\left(\vphantom{\partial \,/\, \partial y}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(y\)\(\left.\vphantom{\partial \,/\, \partial y}\right)\)\(\,-\,\)\(y\)\(\left(\vphantom{\partial \,/\, \partial x}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(x\)\(\left.\vphantom{\partial \,/\, \partial x}\right)\).
However, a relation analogous to CommAngOrb is valid for spins too, though there is nothing like RepAngOrb , namely:
Commutation_angular_spin:
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{S \vphantom{ S}_{x} ,\, S \vphantom{ S}_{y}}\right.\)\(S \vphantom{ S}_{x}\)\(,\,\)\(S \vphantom{ S}_{y}\)\(\left.\vphantom{S \vphantom{ S}_{x} ,\, S \vphantom{ S}_{y}}\right]\)\(\;=\;\)\(S \vphantom{ S}_{z}\)\(;\,\)
As spin is not a classical concept, I take that as an axiom too.
Furthermore, we will need this fact in solving the hydrogen atom:
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{L \vphantom{ L}^{2}_{x} \,+\, L \vphantom{ L}^{2}_{y} \,+\, L \vphantom{ L}^{2}_{y} ,\, L \vphantom{ L}_{z}}\right.\)\(L \vphantom{ L}^{2}_{x}\)\(\,+\,\)\(L \vphantom{ L}^{2}_{y}\)\(\,+\,\)\(L \vphantom{ L}^{2}_{y}\)\(,\,\)\(L \vphantom{ L}_{z}\)\(\left.\vphantom{L \vphantom{ L}^{2}_{x} \,+\, L \vphantom{ L}^{2}_{y} \,+\, L \vphantom{ L}^{2}_{y} ,\, L \vphantom{ L}_{z}}\right]\)\(\;=\;\)\(0\)\(;\,\)
To show it, repeatedly apply CommAngOrb to get \(\,\mathrm{Comm}\,\)\(\left[\vphantom{L \vphantom{ L}^{2}_{x} ,\, L \vphantom{ L}_{z}}\right.\)\(L \vphantom{ L}^{2}_{x}\)\(,\,\)\(L \vphantom{ L}_{z}\)\(\left.\vphantom{L \vphantom{ L}^{2}_{x} ,\, L \vphantom{ L}_{z}}\right]\)\(\,+\,\)\(\,\mathrm{Comm}\,\)\(\left[\vphantom{L \vphantom{ L}^{2}_{y} ,\, L \vphantom{ L}_{z}}\right.\)\(L \vphantom{ L}^{2}_{y}\)\(,\,\)\(L \vphantom{ L}_{z}\)\(\left.\vphantom{L \vphantom{ L}^{2}_{y} ,\, L \vphantom{ L}_{z}}\right]\)\(\,+\,\)\(\,\mathrm{Comm}\,\)\(\left[\vphantom{L \vphantom{ L}^{2}_{z} ,\, L \vphantom{ L}_{z}}\right.\)\(L \vphantom{ L}^{2}_{z}\)\(,\,\)\(L \vphantom{ L}_{z}\)\(\left.\vphantom{L \vphantom{ L}^{2}_{z} ,\, L \vphantom{ L}_{z}}\right]\)\(\;=\;\)\(L \vphantom{ L}^{2}_{x}\)\(L \vphantom{ L}_{z}\)\(\,-\,\)\(\left(\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{z} \,+\, \mathring{\imath} \hslash L \vphantom{ L}_{y}}\right.\)\(L \vphantom{ L}_{x}\)\(L \vphantom{ L}_{z}\)\(\,+\,\)\(\mathring{\imath}\)\(\hslash\)\(L \vphantom{ L}_{y}\)\(\left.\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{z} \,+\, \mathring{\imath} \hslash L \vphantom{ L}_{y}}\right)\)\(L \vphantom{ L}_{x}\)\(\,+\,\)\(L \vphantom{ L}^{2}_{y}\)\(L \vphantom{ L}_{z}\)\(\,-\,\)\(\left(\vphantom{L \vphantom{ L}_{y} L \vphantom{ L}_{z} \,-\, \mathring{\imath} \hslash L \vphantom{ L}_{x}}\right.\)\(L \vphantom{ L}_{y}\)\(L \vphantom{ L}_{z}\)\(\,-\,\)\(\mathring{\imath}\)\(\hslash\)\(L \vphantom{ L}_{x}\)\(\left.\vphantom{L \vphantom{ L}_{y} L \vphantom{ L}_{z} \,-\, \mathring{\imath} \hslash L \vphantom{ L}_{x}}\right)\)\(L \vphantom{ L}_{y}\)\(\;=\;\)\(L \vphantom{ L}_{x}\)\(\left(\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{z} \,-\, L \vphantom{ L}_{z} L \vphantom{ L}_{x}}\right.\)\(L \vphantom{ L}_{x}\)\(L \vphantom{ L}_{z}\)\(\,-\,\)\(L \vphantom{ L}_{z}\)\(L \vphantom{ L}_{x}\)\(\left.\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{z} \,-\, L \vphantom{ L}_{z} L \vphantom{ L}_{x}}\right)\)\(\,+\,\)\(L \vphantom{ L}_{y}\)\(\left(\vphantom{L \vphantom{ L}_{y} L \vphantom{ L}_{z} \,-\, L \vphantom{ L}_{z} L \vphantom{ L}_{y}}\right.\)\(L \vphantom{ L}_{y}\)\(L \vphantom{ L}_{z}\)\(\,-\,\)\(L \vphantom{ L}_{z}\)\(L \vphantom{ L}_{y}\)\(\left.\vphantom{L \vphantom{ L}_{y} L \vphantom{ L}_{z} \,-\, L \vphantom{ L}_{z} L \vphantom{ L}_{y}}\right)\)\(\,+\,\)\(\mathring{\imath}\)\(\hslash\)\(\left(\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{y} \,-\, L \vphantom{ L}_{y} L \vphantom{ L}_{x}}\right.\)\(L \vphantom{ L}_{x}\)\(L \vphantom{ L}_{y}\)\(\,-\,\)\(L \vphantom{ L}_{y}\)\(L \vphantom{ L}_{x}\)\(\left.\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{y} \,-\, L \vphantom{ L}_{y} L \vphantom{ L}_{x}}\right)\)\(\;=\;\)\(\,-\,\)\(\mathring{\imath}\)\(\hslash\)\(L \vphantom{ L}_{x}\)\(L \vphantom{ L}_{y}\)\(\,+\,\)\(\mathring{\imath}\)\(\hslash\)\(L \vphantom{ L}_{y}\)\(L \vphantom{ L}_{x}\)\(\,+\,\)\(\mathring{\imath}\)\(\hslash\)\(\left(\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{y} \,-\, L \vphantom{ L}_{y} L \vphantom{ L}_{x}}\right.\)\(L \vphantom{ L}_{x}\)\(L \vphantom{ L}_{y}\)\(\,-\,\)\(L \vphantom{ L}_{y}\)\(L \vphantom{ L}_{x}\)\(\left.\vphantom{L \vphantom{ L}_{x} L \vphantom{ L}_{y} \,-\, L \vphantom{ L}_{y} L \vphantom{ L}_{x}}\right)\)\(\;=\;\)\(0\).
This indicates that \(L \vphantom{ L}^{2}\) and \(L \vphantom{ L}_{z}\) can be simultaneously diagonalized.
❧ August 27, 2021