Title: Part IB / Quantum Mechanics
Tags: Simultaneous_Measurements Foundations_Of_Quantum_Mechanics 3D-Schrödinger-Equation

Q: Theorem. If \(\psi_1(\mathbf{x}, t)\) and \(\psi_2 (\mathbf{x},t)\) are both normalisable, \[\int_{\mathbb{R}^3} |\psi_1(\mathbf{x}, t)|^2 dV = N_1 < \infty\], \[\int_{\mathbb{R}^3} |\psi_2(\mathbf{x},t)|^2 dV = N_2 < \infty\]then their linear combination is also normalisable.
A: From the triangle inequality we have \[|z_1 + z_2| \leq |z_1| + |z_2|\]and also \[(|z_1| - |z_2|)^2 \geq 0 \implies 2|z_1||z_2| \leq |z_1|^2 + |z_2|^2\]So if we apply these to \(z_1 = a_1 \psi_1, z_2 = a_2 \psi_2\) we get the result by working through the integrals.

Q: Definition. In quantum mechanics, the inner product of two wavefunctions \(\psi, \phi\) at a time t is given by [...]
A: \[(\psi, \phi) = \int_{\mathbb{R}^2} \psi^*(\mathbf{x},t) \phi(\mathbf{x}, t) dV \]
Back Extra: The integral Schwartz inequality tells us that if two wavefunctions are both square integrable, their inner product is guaranteed to exist.

Q: Proposition. Born's Rule. If \(\phi(\mathbf{x},t)\) is the desired outcome of a measurement, then the probability of measuring such an outcome given the wavefunction \(\psi(\mathbf{x},t)\) at a time t is given by \[[...]\]
A: |(\psi, \phi)^2| = \left | \int_{\mathbb{R}^3} \psi^* (\mathbf{x},t) \phi(\mathbf{x}, t) dV \right |^2
Back Extra: We are essentially measuring the 'overlap' of \(\phi\) onto \(\psi\) at time t.

Q: Definition. The particular class of solutions that have the form \[[...]\]where \( \chi (x) \) is an eigenfunction of the Hamiltonian with eigenvalue E, are called stationary states. Stationary states are states of definite energy E.
A: \Psi(x,t) = \chi(x) \exp (\frac{-i}{\hbar} Et)
Back Extra: The probability desnsity is time independent for these states.

Q: Definition. The commutator operator of two operators is defined as \[[...]}\]
A: [\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A
Back Extra: Depends linearly on both entries. It gives a way of calibrating how close two operators are to having simultaneously determinable eigenvalues.

Q: Definition. We say two Hermitian operators \(\hat{A}, \hat{B}\) are simultaneously diagonisable if [...]
A: the space of normalisable wavefunctions has a complete basis of joint eigenfunctions {\(\psi_i\)}
Back Extra: Eigenfunctions such that \(\hat{A}\psi_i = a_i \psi_i, \hat{B} \psi_i = b_i \psi_i\) for some real numbers \(a_i, b_i \in \mathbb{R}\)

Q: Theorem. Two Hermitian operators \(\hat{A}\) and \(\hat{B}\) are simultaneously diagonisable if and only if \([\hat{A}, \hat{B}] = 0\)
A: If they are simultaneously diagonisable, the space of normalisable wavefunctions has a complete basis of joint eigenfunction, now for any such eigenfunction \[[\hat{A}, \hat{B}]\psi_i = \hat{A}\hat{B} \psi_i - \hat{B}\hat{A} \psi_i = (a_i b_i - b_i a_i) \psi_i = 0\]But now any wavefunction can be written as a linear combination of these eigenfunctions so \[[\hat{A}, \hat{B}] \psi = 0\] for all wavefunctions so the commutator operator is the zero operator.
If \([\hat{A}, \hat{B}] = 0\) and \(\hat{A} \psi_i = a_i \psi_i\) then \[0 = \hat{A}(\hat{B} \psi_i) - a_i(\hat{B} \psi_i) \implies \hat{A}(\hat{B} \psi_i) = a_i (\hat{B}) \psi_i\]so \(\hat{B}\psi_i\) is also an eigenfunction of \(\hat{A}\) with eigenvalue \(a_i\). Thus \(\hat{B}\) maps the eigenspace E of \(\hat{A}\) with eigenvalue \(a_i\) to itself, then \(\hat{B}|_E\), the operator \(\hat{B}\) restricted to E is a Hermiatian operator of E. Thus we can find a basis of E in which \(\hat{B}\) acts diagonally, since this is true for all eigenspaces of \(\hat{A}\) we can find a complete basis of simultaneous eigenfunctions of the two operators.

Q: Lemma. The uncertainty \((\Delta_{\psi} A) \geq 0 \) and \((\Delta_{\psi}A) = 0\) if and only if \(\psi\) is an eigenfunction of \(\hat{A}\)
A: \[(\Delta_{\psi}A)^2 = \langle (\hat{A} - \langle \hat{A} \rangle_{\psi})^2 \rangle_{\psi} = ((\hat{A} - \langle \hat{A} \rangle_{\psi}) \psi, (\hat{A} - \langle \hat{A} \rangle_{\psi}) \psi)\]If we now call \(\phi = (\hat{A} - \langle \hat{A} \rangle_{\psi})\psi\), we have \((\phi, \phi) \geq 0 \), with equality only if \(\phi = 0\), i.e. \[\hat{A} \psi = \langle \hat{A} \rangle_{\psi} \psi\]which implies that \(\psi\) is an eigenfuction of \(\hat{A}\).  Conversely, if \(\psi\) is an eigenfunction of \(\hat{A}\) then \(\langle \hat{A} \rangle_{\psi} = (\psi, \hat{A} \psi) = a(\psi, \psi) = a\) and \(\langle \hat{A}^2 \rangle_{\psi} = (\psi, \hat{A}^2 \psi) = a^2 (\psi, \psi) = a^2\), so \(\Delta_{\psi} A = 0\).

Q: Theorem. Schwarz's Inequality. If \(\psi, \phi\) are two normalisable wavefunctions, then \(|(\phi, \psi)|^2 \leq (\phi, \phi)(\psi, \psi)\). We have equality if and only if \(\phi = a \psi\) some complex number a.
A: For any a we have \[0 \leq (\phi - a \psi, \phi - a \psi)\]Take \(a = (\psi, \phi)/(\psi, \psi)\) we have \[0 \leq (\phi, \phi) - 2 \frac{|(\psi, \phi)|^2}{(\psi, \psi)} + \frac{|(\psi, \phi)|^2}{(\psi, \psi)} \implies |(\phi, \psi)|^2 \leq (\phi, \phi) (\psi, \psi)\]with \[|(\phi, \psi)|^2 = (\phi, \phi)(\psi, \psi) \iff \phi = a \psi\] because the norm of a wavefunction is zero only if the wavefunction is 0.

Q: Theorem. Generalised Uncertainty Theorem. If A and B are any two observables, and \(\psi\) is any state, then \[(\Delta_{\psi}A)(\Delta_{\psi}B) \geq \frac{1}{2} |(\psi, [\hat{A}, \hat{B}] \psi)|\]
A: If we write \(\hat{A}' = \hat{A} - \langle \hat{A} \rangle_{\psi}\) and similarly for \(\hat{B}'\) we have \[(\Delta_{\psi}A)^2 (\Delta_{\psi}B)^2 = (\hat{A}' \psi, \hat{A}'\psi)(\hat{B}' \psi, \hat{B}' \psi) \geq |(\hat{A}' \psi, \hat{B}' \psi)|^2 = |(\psi, \hat{A}' \hat{B}' \psi)|^2\]using Schwarz inequality and the fact that \(\hat{A}'\) is Hermitian. We now write \[\hat{A}' \hat{B}' = \frac{1}{2}([\hat{A}', \hat{B}'] + \{\hat{A}', \hat{B}'\})\]where we use the curly brackets for the anti-commutator. Using the defintion of Hermitatian conjugation we can see that \[(\psi, \{\hat{A}', \hat{B}'\}\psi) = (\{\hat{A}', \hat{B}'\}^{\dagger} \psi, \psi) = (\{\hat{A}', \hat{B}' \}\psi, \psi) = (\psi, \{\hat{A}', \hat{B}'\} \psi)^* \]and we get \[(\psi, [\hat{A}' , \hat{B}'] \psi) = - (\psi,[\hat{A}', \hat{B}']\psi )^*\]So \[|(\psi, \hat{A}' \hat{B}' \psi)|^2 = \frac{1}{4}(|(\psi, [\hat{A}', \hat{B}'] \psi)|^2 + |(\psi, \{\hat{A}' , \hat{B}'\} \psi)|^2)\]as the parts are purely real and purely imaginary. Using this we get \[(\Delta_{\psi} A)^2 (\Delta_{\psi} B)^2 \geq \frac{1}{4}|(\psi, [\hat{A}, \hat{B}] \psi)|^2\]and taking the square root we get the theorem.

Q: Lemma. If \[\hat{x} \psi = ia \hat{p}\psi\] for some real parameter a, then \[\Delta_{\psi} \hat{x} \Delta_{\psi} \hat{p} = \hbar / 2\]
A: We have \[(\psi, \{\hat{x}, \hat{p}\} \psi) = (\psi, \hat{x} \hat{p} \psi) + (\psi, \hat{p} \hat{x} \psi) = (\hat{x} \psi, \hat{p} \psi) + (\hat{p} \psi, \hat{x} \psi)= (ia \hat{p} \psi, \hat{p} \psi) + (\hat{p} \psi, ia \hat{p} \psi) = 0\]which is the term we remove in the uncertainty theorem inequality, so we end up with equality.

Q: Theorem. Ehrenfest's Theorem. The expectation value \(\langle \hat{A} \rangle_{\psi}\) of an operator \(\hat{A}\) on a state \(\psi\) evolves by \[\frac{d}{dt} \langle \hat{A} \rangle_{\psi} = \frac{i}{\hbar} \langle [\hat{H}, \hat{A}] \rangle_{\psi} + \langle \frac{\partial \hat{A}}{\partial t} \rangle_{\psi}\]
A: In 1D, omitting the explicit dependence of the wavefunction \(\psi(x,t)\) on x and t, we have \[\frac{d}{dt} \langle \hat{A} \rangle_{\psi} = \frac{d}{dt} \int_{- \infty}^{\infty} \psi^* \hat{A} \psi dx = \int_{- \infty}^{\infty} \left ( \frac{\partial \psi^*}{\partial t} \hat{A} \psi + \psi^* \frac{\partial \hat{A}}{\partial t} \psi + \psi^* \hat{A} \frac{\partial \psi}{\partial t}\right ) dx\]which we simplify to get our result using the Schrödinger equation.

Q: Definition. In Quantum Mechanics, the orbital angular momentum is an observable which corresponds to the operator, \[[...]\]
A: \mathbf{\hat{L} = \hat{x} \times \hat{p} } = -i\hbar\mathbf{x} \times \nabla
Back Extra: We have \(\hat{L}_i = -i \hbar \epsilon_{ijk} x_j \frac{\partial}{\partial x_k}\), and this is Hermitian. THe different components of the angular momentum operator do not commute with each other so we can measure the different components simultaneously.

Q: Definition. In Quantum Mechanics, we define the total angular momentum operator, \[[...]\]
A: \hat{L}^2 = \hat{L}_1^2 + \hat{L}_2^2 + \hat{L}_3^2
Back Extra: The total angular momentum commutes with each of the components of the angular momentum.
The total angular momentum operator commutes with a Hamiltonian featuring a spherically symmetric potential. This means that the angular momentum operators care northing for how the wavefunctions depend on the radial coordinate r.