Students work in small groups to use completeness relations to change the basis of quantum states.

Students use a completeness relations to write hydrogen atoms states in the energy and position bases.

Students use completeness relations to write a matrix element of a spin component in a different basis.

Students practice using inner products to find the components of the cartesian basis vectors in the polar basis and vice versa. Then, students use a completeness relation to change bases or cartesian/polar bases and for different spin bases.

Students use the completeness relation for the position basis to re-express expressions in bra/ket notation in wavefunction notation.

In this lecture, the instructor guides a discussion about translating between bra-ket notation and wavefunction notation for quantum systems.

Students consider the dimensions of spin-state kets and position-basis kets.

We will show that the components of the angular momentum operator \(\vec{L}\), written in differential operator form in rectangular components, satisfy the commutation relations: \begin{equation} \left[L_x,L_y\right]=+i\hbar L_z \qquad(\text{and cyclic permutations}) \end{equation}

First calculate the components of angular momentum classically: \begin{align} \vec{L}&=\vec{r}\times\vec{p}\\ &=(x\hat{x}+y\hat{y}+z\hat{z})\times(p_x\hat{x}+p_y\hat{y}+p_z\hat{z})\\ &=(yp_z-zp_y)\hat{x}+(zp_x-xp_z)\hat{y}+(xp_y-yp_x)\hat{z} \end{align}

Making the standard quantum substitutions, \begin{align} p_x&\rightarrow -i\hbar\partial_x\\ p_y&\rightarrow -i\hbar\partial_y\\ p_z&\rightarrow -i\hbar\partial_z\\ \end{align} we obtain the following operators for the components of angular momentum: \begin{align} \hat{L}_x&=-i\hbar(y\partial_z-z\partial_y)\\ \hat{L}_y&=-i\hbar(z\partial_x-x\partial_z)\\ \hat{L}_z&=-i\hbar(x\partial_y-y\partial_x)\\ \end{align}

To see the role of the product rule in the commutation relations, it is helpful to give the partial derivatives an arbitrary function \(\psi\) to act on. \begin{align} \left[\hat{L}_x,\hat{L}_y\right]\psi &=\left[-i\hbar(y\partial_z-z\partial_y), -i\hbar(z\partial_x-x\partial_z)\right]\psi\\ &=-\hbar^2\left\{(y\partial_z-z\partial_y)(z\partial_x-x\partial_z) -(z\partial_x-x\partial_z)(y\partial_z-z\partial_y)\right\}\psi \end{align} Now, foil-like-mad. Make sure that all of the partial derivatives act on EVERYTHING to their right. Two of the terms above of the form \begin{align} y\,\partial_z(z\,\partial_x \psi) \end{align} require a product rule: \begin{align} y\,\partial_z(z\,\partial_x \psi) &=y((\partial_z z)(\partial_x\psi)+z(\partial_x\partial_x\psi))\\ &=y\partial_x\psi+yz(\partial_x\partial_x\psi) \end{align}

Continuing the calculation above, we see that all of the second derivative terms will cancel because the order of differentiation doesn't matter, leaving only the first derivative terms from the product rule. \begin{align} \left[\hat{L}_x,\hat{L}_y\right]\psi &=\left[-i\hbar(y\partial_z-z\partial_y), -i\hbar(z\partial_x-x\partial_z)\right]\psi\\ &=-\hbar^2\left\{(y\partial_z-z\partial_y)(z\partial_x-x\partial_z) -(z\partial_x-x\partial_z)(y\partial_z-z\partial_y)\right\}\psi\\ &=-\hbar^2\left\{\left(y\,\partial_z(z\,\partial_x \psi) -y\,\partial_z(x\,\partial_z \psi) -z\,\partial_y(z\,\partial_x \psi) +z\,\partial_y(x\,\partial_z \psi)\right)\right.\\ &\;\;\;\quad\quad\left.-\left(z\,\partial_x(y\,\partial_z \psi) -z\,\partial_x(z\,\partial_y \psi) -x\,\partial_z(y\,\partial_z \psi) +x\,\partial_z(z\,\partial_y \psi)\right) \right\}\\ &=-\hbar^2\left\{\left(\cancel{yz(\partial_z\partial_x\psi)} +y\partial_x\psi -\cancel{yx(\partial_z^2\psi)} -\cancel{z^2(\partial_y\partial_x\psi)} +\cancel{zx(\partial_y\partial_z\psi)}\right)\right.\\ &\;\;\;\quad\quad\left.-\left(\cancel{zy(\partial_x\partial_z\psi)} -\cancel{z^2(\partial_x\partial_y\psi)} -\cancel{xy(\partial_z^2\psi)} +\cancel{xz(\partial_z\partial_y\psi)} +x\partial_y\psi\right) \right\}\\ &=i\hbar\left(-i\hbar(-y\partial_x+x\partial_y)\psi\right)\\ &=i\hbar\hat{L}_z\, \psi \end{align} The other components are cyclic permutations of this calculation.

Student explore the properties of an orthonormal basis using the Cartesian and \(S_z\) bases as examples.

• Outer products yield projection operators
• Projection operators are idempotes (they square to themselves)
• A complete set of outer products of an orthonormal basis is the identity (a completeness relation)