1. Quantum mechanics survival guide

1.0. Review of the standard quantum mechanics

In this chapter, we review standard quantum mechanics, highlighting important results. Quantum mechanics describes the evolution of the state of a mechanical system. The state is a vector in a special kind of complex vector space, the Hilbert space. This is a vector space in which an inner product is defined.

We represent the state of the system by a ket-vector, such as $\left| \psi \right\rangle$. The inner product between vectors $\left| \psi\right\rangle$ and $\left| \phi \right\rangle$ is denoted $\left\langle\phi \big| \psi\right\rangle$. For a quantum mechanical system for which we want to calculate the time evolution, the vector describing the system becomes itself time-dependent: $\left| \psi(t)\right\rangle$.

The equation determining the time evolution of a known state vector $\left| \psi(0)\right\rangle$ at $t=0$ is known as the time-dependent Schrödinger equation: $${\rm i}\hslash \frac{\partial}{\partial t} \left| \psi(t) \right\rangle= \hat{H} \left| \psi(t)\right\rangle. \tag{1.1}\label{eq:TDSE11}$$ Here, $\hat{H}$ is an operator acting on vectors in the Hilbert space.

$\hat{H}$ is Hermitian, which means it is equal to its Hermitian conjugate. The Hermitian conjugate $\hat{A}^\dagger$ of an operator $\hat{A}$ is defined as follows: $\hat{A}^\dagger$ must be such that for any two vectors $\psi$ and $\phi$ it must hold that $$\left\langle\phi \right| \hat{A} \left| \psi \right\rangle= \left( \left\langle\psi \right| \hat{A}^\dagger \left|\phi\right\rangle\right)^*,$$

We summarise:

Hermitian operators

$$\hat{A}^\dagger = \hat{A} \text{ means that } \hat{A} \text{ is Hermitian.}$$ In particular, the Hamiltonian $\hat{H}$ is Hermitian: $\hat{H}^\dagger = \hat{H}$.

Hermitian operators have important properties:

• The eigenvalues $\lambda$ of a Hermitian operator are all real: $\lambda = \lambda^*$.

• Eigenvectors $\left| \phi_\lambda \right\rangle$ and $\left| \phi_\mu \right\rangle$ belonging different eigenvalues $\lambda$ and $\mu$ are always mutually orthogonal: $$\left\langle\phi_\lambda \big| \phi_\mu \right\rangle= 0; \phantom{xxx} \lambda \neq \mu.$$

• Degenerate eigenvectors can be chosen orthogonal.

• All eigenvectors span the Hilbert space.

Identity operator

The fact that the eigenvectors form a basis of the Hilbert space leads to the often-used resolution of the identity, in which the unit operator $\unicode{120793}$ is written as: $$\unicode{120793} = \sum_j \left| \phi_\lambda \right\rangle\left\langle \phi_\lambda \right|$$ where the sum is over all the eigenvectors.

Solutions to Schrödinger equation

The solution to the time-dependent Schrödinger equation \eqref{eq:TDSE11} is easy to find: it is just $$\left| \psi(t) \right\rangle= e^{-{\rm i}\hat{H} t/\hslash} \left| \psi(0) \right\rangle.$$ This solution is verified by substituting it back into that equation. However, working out this solution is very difficult, as it contains the exponential of an operator. The easiest way to handle this exponent is by diagonalising the operator. Suppose we have a complete set of eigenvectors $\left| \phi_n \right\rangle$ and eigenvalues $E_n$ for $\hat{H}$: $$\tag{1.2}\label{eq:StatSchrodCh2} \hat{H} \left| \phi_n\right\rangle= E_n \left| \phi_n \right\rangle,$$ and that we know how to expand the initial state $\left| \psi(0)\right\rangle$ into these eigenstates. Given that the exponent of an operator can be written as the diagonal operator with the exponents of the eigenvalues on its diagonal (I write this here for a finite-dimensional matrix): $$e^{-{\rm i}t \hat{H}/\hslash} = \left( \begin{array}{ccccc} e^{-{\rm i}E_1 t/\hslash} & 0 & \cdots & 0 & 0 \\ 0 & e^{-{\rm i}E_2 t/\hslash} & \cdots & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 \\ 0 & 0 & \cdots & 0 & e^{-{\rm i}E_N t/\hslash} \end{array}\right),$$ we obtain $$\left|\psi(t)\right\rangle= \sum_{j=1}^N c_j e^{-{\rm i}E_j t/\hslash} \left| \phi_j\right\rangle,$$ where $$c_j = \left\langle\phi_j | \psi(0)\right\rangle.$$ Eq. \eqref{eq:StatSchrodCh2} is called the stationary Schrödinger equation.

Diagonalising a matrix

In case you have forgotten how to diagonalise a matrix, I recall that, since for an eigenvector $\left|\phi\right\rangle$ of an operator $\hat{A}$, $$\hat{A} \left| \phi \right\rangle= \lambda \left| \phi \right\rangle,$$ $\left| \phi \right\rangle$ should be a non-zero vector for which $$(\hat{A} - \lambda \unicode{120793}) \left|\phi \right\rangle= 0.$$ From your first lecture on linear algebra, you should know that this can only be true if the determinant of the matrix $\hat{A}-\lambda \unicode{120793}$ vanishes. This leads to an algebraic equation for $\lambda$. As an example, consider $$\hat{H} = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right).$$ The determinant condition is $$\left| \begin{array}{cc} -\lambda & 1 \\ 1 & -\lambda \end{array}\right| = 0,$$ which leads to $$\lambda^2 = 1,$$ so $\lambda = \pm 1$. This could be anticipated as you may have recognised $\hat{H}$ as the Pauli matrix $\sigma_x$ and know that the Pauli matrices all have eigenvalues $\pm 1$. Obviously, the larger the matrix, the higher the order of the equation for $\lambda$ and the more work it takes to find the eigenvalues.

Once you have the eigenvalue, you may find the corresponding eigenvector by solving linear equations. Calling the eigenvector $(a,b)$ we have,for the $+1$ eigenvalue, $$\begin{aligned} b & = a \\ a & = b \end{aligned}$$ and the normalised eigenvector becomes $(1,1)/\sqrt{2}$. Similarly we find $(1,-1)/\sqrt{2}$ for the eigenvector with eigenvalue $-1$. These eigenvectors could also have been guessed if you let yourself be guided by the symmetric structure of the matrix you are diagonalising. Another helpful fact is that, for a Hermitian matrix (operator), the eigenvectors belonging to different eigenvalues are orthogonal, and that for equal eigenvalues (degeneracy), all eigenvectors can always be chosen orthogonal. The matrix eigenvalue problem can be solved analytically only if the matrix size is modest (typically smaller than or equal to 3) or if the matrix has a simple and/or very regular structure. In all other cases we use numerical routines for solving the eigenvalue problem.

Time evolution

We see that the time evolution of a wave function is determined by the time evolution operator $\hat{U}(t) = e^{-{\rm i}t \hat{H}/\hslash}$. From the hermiticity of $\hat{H}$, it is easy to see that $\hat{U}(t)$ satisfies $$\hat{U}(t) \hat{U}^\dagger(t) = \hat{U}^\dagger(t) \hat{U}(t) = \unicode{120793},$$ where $\unicode{120793}$ is the unit operator. An operator satisfying this equation is called unitary. We see that unitarity of the time evolution operator directly follows from the hermiticity of the Hamiltonian. Interestingly, this unitarity also guarantees that the norm of the wave function is conserved. To see this, we use $$\begin{aligned} \left| \psi(t) \right\rangle& = \hat{U}(t) \left| \psi(0) \right\rangle; \\ \left\langle\psi(t) \right| & = \left\langle\psi(0) \right| \hat{U}^\dagger(t), \end{aligned}$$ to evaluate what happens to the norm as time evolves: $$\left\langle\psi(t) | \psi(t) \right\rangle = \left\langle\psi(0) \right| \hat{U}^\dagger(t) \hat{U}(t) \left| \psi(0) \right\rangle= \left\langle\psi(0) | \psi(0) \right\rangle,$$ which shows that the norm is indeed preserved.

Example quantum systems with their Hamiltonians

The Schrödinger equation is quite a general equation and does not specify the structure of the Hilbert space, nor the specific form of the Hamiltonian. Physicists have guessed both in the first decades of the twentieth century and good guesses have turned out to yield results for physical measurements in excellent agreement with experiment. Here we list a few.

• Spinless point particle in one dimension. Hilbert space: class of square integrable functions ($L_2$) on the real axis. Hamiltonian: $$\hat{H} = - \frac{\hslash^2}{2m} \frac{d^2}{dx^2} + V(x).$$

• Spinless point particle in three dimensions. Hilbert space: class of square integrable functions ($L_2$) in ${\mathbb R}_3$. Hamiltonian: $$\hat{H} = -\frac{\hslash^2}{2m} \nabla^2 + V({\bf r}).$$

• Particles with spin 1/2, neglecting their motion. Hilbert space: two-dimensional vector space. Hamiltonian: $$\hat{H} = \frac{eB}{m} \sigma_z,$$ where $B$ is a magnetic field along the $z$-axis and $\sigma_z$ is the Pauli matrix $$\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right).$$

It is easy to extend this list with numerous other cases.

Probability and expectation values

For any physical quantity $A$, appropriate for the system at hand, there exists a Hermitian operator $\hat{A}$ whose eigenvalues $\lambda_n$ are the possible values of $A$ found in a measurement. These values occur with probability $\left|\left\langle\phi_n| \psi\right\rangle\right|^2$, where $\phi_n$ is the eigenvector corresponding to $\lambda_n$ and $\left| \psi\right\rangle$ is the state of the system. The expectation value of $A$ in a system in quantum state $\left| \psi \right\rangle$ is given by $\left\langle\psi \right| \hat{A} \left| \psi \right\rangle$.

Radial potential

We now concentrate on electrons in 3D, moving in the field of a radial potential depending only on the distance $r$ to the origin: $V\equiv V(r)$, $r=\left| {\bf r}\right|$. This is a special example of a system exhibiting a symmetry. If there is a symmetry, there usually is degeneracy, meaning that two or more eigenvalues of the Hamiltonian (i.e., the energies) have the same values -- we shall return to this point in chapter 3. It turns out that if there is symmetry, there is one or more operators that commute with the Hamiltonian. The expectation values of these operators (which are assumed to have no explicit time dependence) then remain unchanged in time as can easily be checked from the time evolution: $$\left\langle A \right\rangle_t = \left\langle\psi_0 \left| e^{{\rm i}t H/\hslash} A e^{-{\rm i}t H/\hslash } \right| \psi_0 \right\rangle.$$ Taking the time derivative of this expression yields the commutator $[H,A]$ which vanishes by assumption. It is possible to find a set of vectors that are eigenvectors of all independent operators which commute with $H$. To be specific, states whose energy eigenvalues are degenerate may have different eigenvalues for an operator $\hat{A}$ other than $\hat{H}$. In order to identify all the states uniquely, we need in addition to $\hat{H}$ a set of operators $\hat{A}$, $\hat{B}$, $\hat{C}$,...such that each (simultaneous) eigenvector of this set of operators has a unique set of eigenvalues $E_n, a_j, b_k, c_l$, .... The set of all independent operators which commutes with $H$, including $H$ is called observation maximum:

Observation maximum

An observation maximum is the set of all independent operators, including H, that commute amongst themselves and with H. The eigenvalues of all these operators label the simultaneous eigenvectors of all the operators of the observation maximum. They form a basis of the Hilbert space.

Degeneracy

Degeneracy is related to symmetry which is present in the Hamiltonian (this is formally substantiated by the quantum mechanical version of Noether's theorem, which we shall not go into here). This relation is the reason why the energies of a 3D system which is spherically symmetric (that is, a system with a radial potential), are degenerate. The operators commuting with the Hamiltonian for a spinless particle in a radial potential are the angular momentum operators $L^2$ and $L_z$. These have the eigenvalues $\hslash^2 l(l+1)$ and $\hslash m$ respectively, where $l$ is $0, 1, 2, \ldots$ and $m$ is an integer running from $-l$ to $l$. The energy eigenvalues depend on $l$ and an additional quantum number, $n$ -- they are written as $E_{nl}$. For each $l$, $m$ runs from $-l$ to $l$ in integer steps. So there are $2l+1$ $m$-values for each $l$. As the energy eigenvalues do not depend on $m$, they are (at least) $2l+1$-fold degenerate. If the particles have spin-1/2, there are additional quantum numbers: $s$ which always takes on the value $1/2$ (as we are dealing with spin-1/2 particles) and $m_s$ which takes the value $\pm 1 /2$. All in all, the states of an electron in a radial potential are denoted $\left| n, l, m, s, m_s\right\rangle$. The quantum number $s$ being always 1/2 for an electron, is often left out. If we include the spin, each level with quantum number $l$ is $2(2l+1)$-fold degenerate. Figure 1.1 shows a schematic representation of a spectrum of a particle in a radial potential.

The spectrum of an electron in a radial potential, grouped according to the $l$-value. In parenthesis, the degeneracy of the levels (including the two-fold spin-degeneracy) is given.

Coulomb potential

The Coulomb potential is a special case: this potential has some hidden symmetry which causes several of the $E_{nl}$ to coincide¹. Whereas we normally label the energy eigenvalues for each $l$ by $n=1, 2, \ldots$, this degeneracy allows us to label the states as in the figure below.

The spectrum of an electron in a Coulomb potential, grouped according to their $l$-value. The numbers along the vertical axis is the principal quantum number. The degeneracies per $l$-value are the same as for radial potential (see figure above). Adding the degeneracies for each principal quantum number $n$ gives the degeneracies in parentheses on the right hand side.

The quantum number along the vertical axis is called the principal quantum number. We see that for each principal quantum number $n$, we have states with $l$-values between $0$ and $n-1$. This is a special degeneracy of the hydrogen atom -- the degeneracies of the energy levels can be found by adding the degeneracies for each $l=0, \ldots, n-1$. These degeneracies correspond to the so-called 'noble gas' atoms (for higher $n$, deviations from this series occur due to effects not taken into account here).

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1. This symmetry is related to the four-dimensional rotational group $O(4)$. ↩