Quantum Mechanics

Momentum operator \(\hat{\mathbf{p}}\) is the generator of translation \(\hat{U}(\mathbf{x})\): \[ \hat{U}(\mathbf{x}) = e^{-i\hat{\mathbf{p}}\vdot \mathbf{x}/\hbar}. \]

\begin{align*} \grad \hat{U}(\mathbf{x})\ket{\mathbf{x}} &= \left( \frac{\hat{U}(\dd{\mathbf{x}}) - \hat{U}(\mathbf{0})}{\dd{\mathbf{x}}} \right)\ket{\mathbf{x}} \\ &= \frac{\ket{\mathbf{x} + \dd{\mathbf{x}}} - \ket{\mathbf{x}}}{\dd{\mathbf{x}}} \\ &= \pdv{\mathbf{x}} \ket{\mathbf{x}} \\ &= \frac{1}{i\hbar} \hat{\mathbf{p}} \ket{\mathbf{x}}. \end{align*}

The position basis form of momentum operator is derived as follows:

\begin{align*} \mel{\mathbf{x}}{\hat{\mathbf{p}}}{\psi} &= \bra{\mathbf{x}} \int_{\Omega} \dd[3]{\mathbf{x}'} \psi(\mathbf{x}')\hat{\mathbf{p}}\ket{\mathbf{x}'} \\ &= \bra{\mathbf{x}} \int_{\Omega} \dd[3]{\mathbf{x}'} \psi(\mathbf{x}')i\hbar \pdv{\mathbf{x}'}\ket{\mathbf{x}'} \\ &= \bra{\mathbf{x}} \int_{\Omega} \dd[3]{\mathbf{x}'}i\hbar \pdv{\mathbf{x}'}\left( \psi(\mathbf{x}')\ket{\mathbf{x}'} \right) + \bra{\mathbf{x}} \int_{\Omega} \dd[3]{\mathbf{x}'} \left( - i\hbar \pdv{\mathbf{x}'}\psi(\mathbf{x}') \right)\ket{\mathbf{x}'}\\ &= \bra{\mathbf{x}}\oint_{\partial \Omega}\dd[2]{\mathbf{x}'}i\hbar \psi(\mathbf{x}')\ket{\mathbf{x}'} + \bra{\mathbf{x}} \int_{\Omega} \dd[3]{\mathbf{x}'} \left( - i\hbar \pdv{\mathbf{x}'}\psi(\mathbf{x}') \right)\ket{\mathbf{x}'} \\ &= \bra{\mathbf{x}} \int_{\Omega} \dd[3]{\mathbf{x}'} \left( - i\hbar \pdv{\mathbf{x}'}\psi(\mathbf{x}') \right)\ket{\mathbf{x}'} \\ &= \int_{\Omega} \dd[3]{\mathbf{x}'} \left( - i\hbar \pdv{\mathbf{x}'}\psi(\mathbf{x}') \right)\delta^3(\mathbf{x} - \mathbf{x}') \\ &= -i\hbar \pdv{\mathbf{x}} \psi(\mathbf{x}) \end{align*}

where the wave function \(\psi(\mathbf{x})\) is assumed to be localized and the boundary term vanishes.

The unitary operators \(\hat{U}\) form a Lie group, and the Hermitian operators \(\hat{H}\) are related to the Lie algebra that generate that Lie group by the exponential map: \[ \exp\colon \frac{1}{i\hbar}\hat{H} \mapsto \hat{U}. \]

Hermitian operator of an observable \(p\) maps to transformation in its conjugate variable \(q\). The justification comes from the quantum equivalent of Hamilton's equations:

\begin{align*} &\pdv{q}\ev{H} = - \dv{t} \ev{p} \\ \implies &\pdv{\bra{\psi}}{q} \hat{H}\ket{\psi} + \bra{\psi} \pdv{\hat{H}}{q}\ket{\psi} + \bra{\psi}\hat{H} \pdv{\ket{\psi}}{q}= - \left(\dv{\bra{\psi}}{t} \hat{p}\ket{\psi} + \bra{\psi} \dv{\hat{p}}{t}\ket{\psi} + \bra{\psi}\hat{p} \dv{\ket{\psi}}{t}\right). \end{align*}

Let us assume that the Hamiltonian operator does not explicitly depend on \(q\) (not \(\hat{q}\)), and canonical momentum operator \(\hat{p}\) does not explicitly depend on time. Using Schrödinger Equation, the equation becomes:

\begin{align*} \pdv{\bra{\psi}}{q} \hat{H}\ket{\psi} + &\bra{\psi}\hat{H} \pdv{\ket{\psi}}{q} \\ = \mel{\psi}{-\frac{1}{i\hbar}\hat{p}\hat{H}}{\psi} + &\mel{\psi}{\hat{H}\frac{1}{i\hbar}\hat{p}}{\psi}. \end{align*}

This strongly suggests the following identity \[ i\hbar \pdv{q} \ket{\psi} = \hat{p} \ket{\psi} \] and consequently the generator relation: \[ \hat{U}(q) = \exp(\frac{\hat{p} q}{i\hbar}). \] where \(\hat{U}(q)\) is the translation operator in the general coordinate \(q\) by \(q\) amount.

Similar argument can be made for the second Hamilton equation. From \[ \pdv{p}\ev{H} = \dv{t}\ev{q}, \] we get: \[ -i\hbar \pdv{p}\ket{\psi} = \hat{q}\ket{\psi} \] and \[ \hat{U}(p) = \exp(-\frac{\hat{q}p}{i\hbar}) \]

• Starts positive at \(r=0\) , flips sign \(n-l-1\) times.
• flattens as \(n\) gets large.

• basis function for all sperical functions.
• smooth at every points.
• it is the orbitals in chemistry when the real part of the function is taken.
• it flips its phase \(l-|m|\) times, when going from north pole to the south.
• it rotates in phase \(m\) times, as circling along the azimuthal plane.

\begin{align*} \sigma_1 = \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ \sigma_2 = \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \\ \sigma_3 = \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ \end{align*}

Sometimes the identity matrix is included as \(\sigma_0\).

• Hermitian, Involutory, Unitary.
• They, together with the identity matrix \(I = \sigma_0\), form the basis for the real vector space of \(2\times 2\) Hermitian matrices.
• Anti-Commutativity under Multiplication.
• It is the linear map (order 3 tensor) from the three dimensional vector space \(V\) to the spinor-pair space \(S\otimes S^*\):
  • \[ \sigma_i\vphantom{\sigma}^{a}\vphantom{\sigma}_b \]
  • where \(a\) is the index for the spinor space \(S\) and \(b\) is the index for the dual spinor space \(S^*\).
• It is with respect to the standard spinor basis.

• It represents the interaction between the spin of a particle with an external electromagnetic field, which is best represented in the
  • \(\sigma s = s'\)
• It also represents the observables which are the Hermitian matrices.
  • \(\langle s | \sigma | s \rangle\)
• It is equivalent to the three basis vector of the Clifford Algebra \(\mathrm{Cl}_{3,0}(\mathbb{R})\).

\[ \sigma_\mu^{a\dot{b}} \]

• Generalization of the Pauli matrices to the four dimensional spacetime.
• Using the 5.5.3.1.

• Conjugation rotates \(\pi\) radians around the axis:

  \begin{align*} \sigma_x &\rightsquigarrow \sigma_z\sigma_x\sigma_z^{-1} = \sigma_z\sigma_x\sigma_z = -\sigma_z\sigma_z\sigma_x = -\sigma_x \\ \sigma_y &\rightsquigarrow \sigma_z\sigma_y\sigma_z^{-1} = \sigma_z\sigma_y\sigma_z = -\sigma_z\sigma_z\sigma_y = -\sigma_y \\ \sigma_z &\rightsquigarrow \sigma_z\sigma_z\sigma_z^{-1} = \sigma_z\sigma_z\sigma_z = +\sigma_z \end{align*}
• And negative conjugation by \(\sigma_z\) reflects along the \(z\) axis.

• Two reflections is a rotation.
• Pauli vectors are reflected by the Pauli matrices by the conjugation.
• And rotated by the two pairs of unit Pauli vectors, which are the members of the \(\mathrm{SU}(2)\) group.
• Using some abstract algebra, the \(\mathrm{SU}(2)\) matrices are all the matrix the rotates a Pauli vector, (up to a complex multiple, which does not matter what that is so the determinant is just taken to be 1).

• Traceless Hermitian Matrix
• It squares to the square magnitude of the vector.
• The determinant is the negative of the square magnitude.
• The sum of pairs of sigma matrices are equivalent to the quaternion.
• Both the group of unit quaternions and \(\mathrm{SU}(2)\) are isomorphic to the \(\mathrm{Spin}(3)\) Group, describing the spin in three dimensional space, which is the double cover of rotation group \(\mathrm{SO}(3)\).

• It is rotated by a single \(\mathrm{SU}(2)\) matrix, which is half of the rotation of the Pauli vector, that does not change the length of the spinor: \[ \lVert U\xi \rVert^2 = (U\xi)^\dagger (U\xi) = \xi^\dagger U^\dagger U\xi = \xi^\dagger \xi = \lVert \xi \rVert^2. \]
• The spinors on the Poincaré sphere and the Bloch sphere is also rotated in the same way.

• Hermitian matrix
• The determinant is the spacetime interval.
• It transforms with \(\mathrm{SL}(2, \mathbb{C})\) matrices.
  • The transformation matrix is special and Hermitian
  • The boosts uses the hyperbolic functions

• \(\psi^\dagger \epsilon \phi\) is never conserved.
• So \(\psi^\top \epsilon \phi\) is chosen as the inner product with
  • \[ \epsilon = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}. \]
• It is alternating.
• \(\epsilon\) is called a

• From the inner product

\[ \psi^\top\epsilon \]

• The complex conjugate of a left spinor \(\psi\) becomes the right dual spinor \(\psi^*\).
• The right spinor would be \(\psi^\dagger \epsilon\).

• It has left and right chirality, which is notated by \(\psi^a, \psi_a\) and \(\psi^{\dot{a}}, \psi_{\dot{a}}\).
• This happens because the \(\mathrm{SO}(2,\mathbb{C})\) has two different representation, left and right. The two are connected by complex conjugation.
  • Unlike \(\mathrm{SU}(2)\), there is no invertible matrix \(P\), such that \(L^* = P^{-1}LP\).