# College Math Teaching

## August 19, 2011

### Partial Differential Equations, Differential Equations and the Eigenvalue/Eigenfunction problem

Suppose we are trying to solve the following partial differential equation:
$\frac{\partial \psi}{\partial t} = 3 \frac{\partial ^2 \phi}{\partial x^2}$ subject to boundary conditions:
$\psi(0) = \psi(\pi) = 0, \psi(x,0) = x(x-\pi)$

It turns out that we will be using techniques from ordinary differential equations and concepts from linear algebra; these might be confusing at first.

The first thing to note is that this differential equation (the so-called heat equation) is known to satisfy a “uniqueness property” in that if one obtains a solution that meets the boundary criteria, the solution is unique. Hence we can attempt to find a solution in any way we choose; if we find it, we don’t have to wonder if there is another one lurking out there.

So one technique that is often useful is to try: let $\psi = XT$ where $X$ is a function of $x$ alone and $T$ is a function of $t$ alone. Then when we substitute into the partial differential equation we obtain:
$XT^{\prime} = 3X^{\prime\prime}T$ which leads to $\frac{T^{\prime}}{T} = 3\frac{X^{\prime\prime}}{X}$

The next step is to note that the left hand side does NOT depend on $x$; it is a function of $t$ alone. The right hand side does not depend on $t$ as it is a function of $x$ alone. But the two sides are equal; hence neither side can depend on $x$ or $t$; they must be constant.

Hence we have $\frac{T^{\prime}}{T} = 3\frac{X^{\prime\prime}}{X} = \lambda$

So far, so good. But then you are told that $\lambda$ is an eigenvalue. What is that about?

The thing to notice is that $T^{\prime} - \lambda T = 0$ and $X^{\prime\prime} - \frac{\lambda}{3}X = 0$
First, the equation in $T$ can be written as $D(T) = \lambda T$ with the operator $D$ denoting the first derivative. Then the second can be written as $D^2(X) = 3\lambda X$ where $D^2$ denotes the second derivative operator. Recall from linear algebra that these operators meet the requirements for a linear transformation if the vector space is the set of all functions that are “differentiable enough”. So what we are doing, in effect, are trying to find eigenvectors for these operators.

So in this sense, solving a homogeneous differential equation is really solving an eigenvector problem; often this is termed the “eigenfucntion” problem.

Note that the differential equations are not difficult to solve:
$T = a exp(\lambda T)$ $X = b exp(\sqrt{\frac{\lambda}{3}} x) + cexp(-\sqrt{\frac{\lambda}{3}} x)$; the real valued form of the equation in $x$ depends on whether $\lambda$ is positive, zero or negative.

But the point is that we are merely solving a constant coefficient differential equation just as we did in our elementary differential equations course with one important difference: we don’t know what the constant (the eigenvalue) is.

Now if we turn to the boundary conditions on $x$ we see that a solution of the form $A e^{bx} + Be^{-bx}$ cannot meet the zero at the boundaries conditions; we can rule out the $\lambda = 0$ condition as well.
Hence we know that $\lambda$ is negative and we get $X = a cos(\sqrt{\frac{\lambda}{3}} x) + b sin(\sqrt{\frac{\lambda}{3}} x)$ solution and then $T = d e^{\lambda t }$ solution.

But now we notice that these solutions have a $\lambda$ in them; this is what makes these ordinary differential equations into an “eigenvalue/eigenfucntion” problem.

So what values of $\lambda$ will work? We know it is negative so we say $\lambda = -w^2$ If we look at the end conditions and note that $T$ is never zero, we see that the cosine term must vanish ($a = 0$ ) and we can ensure that $\sqrt{\frac{w}{3}}\pi = k \pi$ which implies that $w = 3k^2$ So we get a whole host of functions: $\psi_k = a_k e^{-3k^2 t}sin(kx)$.

Now we still need to meet the last condition (set at $t = 0$ ) and that is where Fourier analysis comes in. Because the equation was linear, we can add the solutions and get another solution; hence the $X$ term is just obtained by taking the Fourier expansion for the function $x(x-\pi)$ in terms of sines.

The coefficients are $b_k = \frac{1}{\pi} \int^{\pi}_{-\pi} (x)(x-\pi) sin(kx) dx$ and the solution is:
$\psi(x,t) = \sum_{k=1}^{\infty} e^{-3k^2 t} b_k sin(kx)$

## August 17, 2011

### Quantum Mechanics and Undergraduate Mathematics XIV: bras, kets and all that (Dirac notation)

Filed under: advanced mathematics, applied mathematics, linear albegra, physics, quantum mechanics, science — collegemathteaching @ 11:29 pm

Up to now, I’ve used mathematical notation for state vectors, inner products and operators. However, physicists use something called “Dirac” notation (“bras” and “kets”) which we will now discuss.

Recall: our vectors are integrable functions $\psi: R^1 \rightarrow C^1$ where $\int^{-\infty}_{\infty} \overline{\psi} \psi dx$ converges.

Our inner product is: $\langle \phi, \psi \rangle = \int^{-\infty}_{\infty} \overline{\phi} \psi dx$

Here is the Dirac notation version of this:
A “ket” can be thought of as the vector $\langle , \psi \rangle$. Of course, there is an easy vector space isomorphism (Hilbert space isomorphism really) between the vector space of state vectors and kets given by $\Theta_k \psi = \langle,\psi \rangle$. The kets are denoted by $|\psi \rangle$.
Similarly there are the “bra” vectors which are “dual” to the “kets”; these are denoted by $\langle \phi |$ and the vector space isomorphism is given by $\Theta_b \psi = \langle,\overline{\psi} |$. I chose this isomorphism because in the bra vector space, $a \langle\alpha,| = \langle \overline{a} \alpha,|$. Then there is a vector space isomorphism between the bras and the kets given by $\langle \psi | \rightarrow |\overline{\psi} \rangle$.

Now $\langle \psi | \phi \rangle$ is the inner product; that is $\langle \psi | \phi \rangle = \int^{\infty}_{-\infty} \overline{\psi}\phi dx$

By convention: if $A$ is a linear operator, $\langle \psi,|A = \langle A(\psi)|$ and $A |\psi \rangle = |A(\psi) \rangle$ Now if $A$ is a Hermitian operator (the ones that correspond to observables are), then there is no ambiguity in writing $\langle \psi | A | \phi \rangle$.

This leads to the following: let $A$ be an operator corresponding to an observable with eigenvectors $\alpha_i$ and eigenvalues $a_i$. Let $\psi$ be a state vector.
Then $\psi = \sum_i \langle \alpha_i|\psi \rangle \alpha_i$ and if $Y$ is a random variable corresponding to the observed value of $A$, then $P(Y = a_k) = |\langle \alpha_k | \psi \rangle |^2$ and the expectation $E(A) = \langle \psi | A | \psi \rangle$.

## August 9, 2011

### Quantum Mechanics and Undergraduate Mathematics IX: Time evolution of an Observable Density Function

We’ll assume a state function $\psi$ and an observable whose Hermitian operator is denoted by $A$ with eigenvectors $\alpha_k$ and eigenvalues $a_k$. If we take an observation (say, at time $t = 0$ ) we obtain the probability density function $p(Y = a_k) = | \langle \alpha_k, \psi \rangle |^2$ (we make the assumption that there is only one eigenvector per eigenvalue).

We saw how the expectation (the expected value of the associated density function) changes with time. What about the time evolution of the density function itself?

Since $\langle \alpha_k, \psi \rangle$ completely determines the density function and because $\psi$ can be expanded as $\psi = \sum_{k=1} \langle \alpha_k, \psi \rangle \alpha_k$ it make sense to determine $\frac{d}{dt} \langle \alpha_k, \psi \rangle$. Note that the eigenvectors $\alpha_k$ and eigenvalues $a_k$ do not change with time and therefore can be regarded as constants.

$\frac{d}{dt} \langle \alpha_k, \psi \rangle = \langle \alpha_k, \frac{\partial}{\partial t}\psi \rangle = \langle \alpha_k, \frac{-i}{\hbar}H\psi \rangle = \frac{-i}{\hbar}\langle \alpha_k, H\psi \rangle$

We can take this further: we now write $H\psi = H\sum_j \langle \alpha_j, \psi \rangle \alpha_j = \sum_j \langle \alpha_j, \psi \rangle H \alpha_j$ We now substitute into the previous equation to obtain:
$\frac{d}{dt} \langle \alpha_k, \psi \rangle = \frac{-i}{\hbar}\langle \alpha_k, \sum_j \langle \alpha_j, \psi \rangle H \alpha_j \rangle = \frac{-i}{\hbar}\sum_j \langle \alpha_k, H\alpha_j \rangle \langle \alpha_j, \psi \rangle$

Denote $\langle \alpha_j, \psi \rangle$ by $a_j$. Then we see that we have the infinite coupled differential equations: $\frac{d}{dt} a_k = \frac{-i}{\hbar} \sum_j a_j \langle \alpha_k, H\alpha_j \rangle$. That is, the rate of change of one of the $a_k$ depends on all of the $a_j$ which really isn’t a surprise.

We can see this another way: because we have a density function, $\sum_j |\langle \alpha_j, \psi \rangle |^2 =1$. Now rewrite: $\sum_j |\langle \alpha_j, \psi \rangle |^2 = \sum_j \langle \alpha_j, \psi \rangle \overline{\langle \alpha_j, \psi \rangle } = \sum_j a_j \overline{ a_j} = 1$. Now differentiate with respect to $t$ and use the product rule: $\sum_j \frac{d}{dt}a_j \overline{ a_j} + a_j \frac{d}{dt} \overline{ a_j} = 0$

Things get a bit easier if the original operator $A$ is compatible with the Hamiltonian $H$; in this case the operators share common eigenvectors. We denote the eigenvectors for $H$ by $\eta$ and then
$\frac{d}{dt} a_k = \frac{-i}{\hbar} \sum_j a_j \langle \alpha_k, H\alpha_j \rangle$ becomes:
$\frac{d}{dt} \langle \eta_j, \psi \rangle = \frac{-i}{\hbar} \sum_j \langle \eta_j, \psi \rangle \langle \eta_k, H\eta_j \rangle$ Now use the fact that the $\eta_j$ are eigenvectors for $H$ and are orthogonal to each other to obtain:
$\frac{d}{dt} \langle \eta_k, \psi \rangle = \frac{-i}{\hbar} e_k \langle \eta_k, \psi \rangle$ where $e_k$ is the eigenvalue for $H$ associated with $\eta_k$.

Now we use differential equations (along with existence and uniqueness conditions) to obtain:
$\langle \eta_k, \psi \rangle = \langle_k, \psi_0 \rangle exp(-ie_k \frac{t}{\hbar})$ where $\psi_0$ is the initial state vector (before it had time to evolve).

This has two immediate consequences:

1. $\psi(x,t) = \sum_j \langle \eta_j, \psi_0 \rangle exp(-ie_j \frac{t}{\hbar}) \eta_j$
That is the general solution to the time-evolution equation. The reader might be reminded that $exp(ib) = cos(b) + i sin (b)$

2. Returning to the probability distribution: $P(Y = e_k) = |\langle \eta_k, \psi \rangle |^2 = |\langle \eta_k, \psi_0 \rangle |^2 ||exp(-ie_k \frac{t}{\hbar})|^2 = |\langle \eta_k, \psi_0 \rangle |^2$. But since $A$ is compatible with $H$, we have the same eigenvectors, hence we see that the probability density function does not change AT ALL. So such an observable really is a “constant of motion”.

Stationary States
Since $H$ is an observable, we can always write $\psi(x,t) = \sum_j \langle \eta_j, \psi(x,t) \rangle \eta_j$. Then we have $\psi(x,t)= \sum_j \langle \eta_j, \psi_0 \rangle exp(-ie_j \frac{t}{\hbar}) \eta_j$

Now suppose $\psi_0$ is precisely one of the eigenvectors for the Hamiltonian; say $\psi_0 = \eta_k$ for some $k$. Then:

1. $\psi_(x,t) = exp(-ie_k \frac{t}{\hbar}) \eta_k$
2. For any $t \geq 0 , P(Y = e_k) = 1, P(Y \neq e_k) = 0$

Note: no other operator has made an appearance.
Now recall our first postulate: states are determined only up to scalar multiples of unity modulus. Hence the state undergoes NO time evolution, no matter what observable is being observed.

We can see this directly: let $A$ be an operator corresponding to any observable. Then $\langle \alpha_k, A \psi_k \rangle = \langle \alpha_k, A exp(-i e_k \frac{t}{\hbar})\eta_k \rangle = exp(-i e_k \frac{t}{\hbar}\langle \alpha_k, A \eta_k \rangle$. Then because the probability distribution is completely determined by the eigenvalues $e_k$ and $|\langle \alpha_k, A \eta_k \rangle |$ and $|exp(-i e_k \frac{t}{\hbar}| = 1$, the distribution does NOT change with time. This motivates us to define the stationary states of a system: $\psi_{(k)} = exp(- e_k \frac{t}{\hbar})\eta_k$.

Gillespie notes that much of the problem solving in quantum mechanics is solving the Eigenvalue problem: $H \eta_k = e_k \eta_k$ which is often difficult to do. But if one can do that, one can determine the stationary states of the system.

## July 25, 2011

### Quantum Mechanics and Undergraduate Mathematics V: compatible observables

This builds on our previous example. We start with a state $\psi$ and we will make three successive observations of observables which have operators $A$ and $B$ in the following order: $A, B, A$. The assumption is that these observations are made so quickly that no time evolution of the state vector can take place; all of the change to the state vector will be due to the effect of the observations.

A simplifying assumption will be that the observation operators have the following property: no two different eigenvectors have the same eigenvalues (e. g., the eigenvalue uniquely determines the eigenvector up to multiplication by a constant of unit modulus).

First of all, this is what “compatible observables” means: two observables $A, B$ are compatible if, upon three successive measurements $A, B, A$ the first measurement of $A$ is guaranteed to be the second measurement of $A$. That is, the state vector after the first measurement of $A$ is the same state vector after the second measurement of $A$.

So here is what the compatibility theorem says (I am freely abusing notation by calling the observable by the name of its associated operator):

Compatibility Theorem
The following are equivalent:

1. $A, B$ are compatible observables.
2. $A, B$ have a common eigenbasis.
3. $A, B$ commute (as operators)

Note: for this discussion, we’ll assume an eigenbasis of $\alpha_i$ for $A$ and $\beta_i$ for $B$.

1 implies 2: Suppose the state of the system is $\alpha_k$ just prior to the first measurement. Then the first measurement is $a_k$. The second measurement yields $b_j$ which means the system is in state $\beta_j$, in which case the third measurement is guaranteed to be $a_k$ (it is never anything else by the compatible observable assumption). Hence the state vector must have been $\alpha_k$ which is the same as $\beta_j$. So, by some reindexing we can assume that $\alpha_1 = \beta_1$. An argument about completeness and orthogonality finishes the proof of this implication.

2 implies 1: after the first measurement, the state of the system is $\alpha_k$ which, being a basis vector for observable $B$ means that the system after the measurement of $B$ stays in the same state, which implies that the state of the system will remain $\alpha_k$ after the second measurement of $A$. Since this is true for all basis vectors, we can extend this to all state vectors, hence the observables are compatible.

2 implies 3: a common eigenbasis implies that the operators commute on basis elements so the result follows (by some routine linear-algebra type calculations)

3 implies 2: given any eigenvector $\alpha_k$ we have $AB \alpha_k = BA \alpha_k = a_k B \alpha_k$ which implies that $B \alpha_k$ is an eigenvector for $A$ with eigenvalue $\alpha_k$. This means that $B \alpha_k = c \alpha_k$ where $c$ has unit modulus; hence $\alpha_k$ must be an eigenvector of $B$. In this way, we establish a correspondence between the eigenbasis of $B$ with the eigenbasis of $A$.

Ok, what happens when the observables are NOT compatible?

Here is a lovely application of conditional probability. It works this way: suppose on the first measurement, $a_k$ is observed. This puts us in state vector $\alpha_k$. Now we measure the observable $B$ which means that there is a probability $|\langle \alpha_k, \beta_i \rangle|^2$ of observing eigenvalue $b_i$. Now $\beta_i$ is the new state vector and when observable $A$ is measured, we have a probability $|\langle \alpha_j, \beta_i \rangle|^2$ of observing eigenvalue $a_j$ in the second measurement of observable $A$.

Therefore given the initial measurement we can construct a conditional probability density function $p(a_j|a_k) = \sum_i p(b_i|a_k)p(a_j|b_i)= \sum_i |\langle \alpha_k, \beta_i \rangle| |^2 |\langle \beta_i, \alpha_j |^2$

Again, this makes sense only if the observations were taken so close together so as to not allow the state vector to undergo time evolution; ONLY the measurements changes the state vector.

Next: we move to the famous Heisenberg Uncertainty Principle, which states that, if we view the interaction of the observables $A$ and $B$ with a set state vector and abuse notation a bit and regard the associated density functions (for the eigenvalues) by the same letters, then $V(A)V(B) \geq (1/4)|\langle \psi, [AB-BA]\psi \rangle |^2.$

Of course, if the observables are compatible, then the right side becomes zero and if $AB-BA = c$ for some non-zero scalar $c$ (that is, $(AB-BA) \psi = c\psi$ for all possible state vectors $\psi$ ), then we get $V(A)V(B) \geq (1/4)|c|^2$ which is how it is often stated.

## July 15, 2011

### Quantum Mechanics and Undergraduate Mathematics III: an example of a state function

I feel bad that I haven’t given a demonstrative example, so I’ll “cheat” a bit and give one:

For the purposes of this example, we’ll set our Hilbert space to the the square integrable piecewise smooth functions on $[-\pi, \pi]$ and let our “state vector” $\psi(x) =\left\{ \begin{array}{c}1/\sqrt{\pi}, 0 < x \leq \pi \\ 0,-\pi \leq x \leq 0 \end{array}\right.$

Now consider a (bogus) state operator $d^2/dx^2$ which has an eigenbasis $(1/\sqrt{\pi})cos(kx), (1/\sqrt{\pi})sin(kx), k \in {, 1, 2, 3,...}$ and $1/\sqrt{2\pi}$ with eigenvalues $0, -1, -4, -9,......$ (note: I know that this is a degenerate case in which some eigenvalues share two eigenfunctions).

Note also that the eigenfunctions are almost the functions used in the usual Fourier expansion; the difference is that I have scaled the functions so that $\int^{\pi}_{-\pi} (sin(kx)/\sqrt{\pi})^2 dx = 1$ as required for an orthonormal basis with this inner product.

Now we can write $\psi = 1/(2 \sqrt{\pi}) + 4/(\pi^{3/2})(sin(x) + (1/3)sin(3x) + (1/5)sin(5x) +..)$
(yes, I am abusing the equal sign here)
This means that $b_0 = 1/\sqrt{2}, b_k = 2/(k \pi), k \in {1,3,5,7...}$

Now the only possible measurements of the operator are 0, -1, -4, -9, …. and the probability density function is: $p(A = 0) = 1/2, P(A = -1) = 4/(\pi^2), P(A = -3) = 4/(9 \pi^2),...P(A = -(2k-1))= 4/(((2k-1)\pi)^2)..$

One can check that $1/2 + (4/(\pi^2))(1 + 1/9 + 1/25 + 1/49 + 1/81....) = 1.$

Here is a plot of the state function (blue line at the top) along with some of the eigenfunctions multiplied by their respective $b_k$.

## July 13, 2011

### Quantum Mechanics and Undergraduate Mathematics II

In the first part of this series, we reviewed some of the mathematical background that we’ll use. Now we get into a bit of the physics.

For simplification, we’ll assume one dimensional, non-relativistic motion. No, nature isn’t that simple; that is why particle physics is hard! 🙂

What we will do is to describe a state of a system and the observables. The state of the system is hard to describe; in the classical case (say the damped mass-spring system in harmonic motion), the state of the system is determined by the system parameters (mass, damping constant, spring constant) and the velocity and acceleration at a set time.

And observable is, roughly speaking, something that can give us information about the state of the system. In classical mechanics, one observable might be $H(x, p) = P^2/2m + V(x)$ where $p$ is the system’s momentum and $V(x)$ represents the potential energy at position $x$. If this seems strange, remember that $p = mv$ therefore kinetic energy is $mv^2/2$ and solving for momentum $p$ gives us the formula. We bring this up because something similar will appear later.

In quantum mechanics, certain postulates are assumed. I’ll present the ones that Gillespie uses:

Postulate 1: Every possible physical state of a given system corresponds to a Hilbert space vector $\psi$ of unit norm (using the inner product that we talked about) and every such vector corresponds to a possible state of a system. The correspondence of states to the vectors is well defined up to multiplication of a vector by a complex number of unit modulus.

Note: this state vector, while containing all of the knowable information of the system, says nothing about what could be known or how such knowledge might be observed. Of course, this state vector might evolve with time and sometimes it is written as $\psi_{t}$ for this reason.

Postulate 2 There is a one to one correspondence between physical observables and linear Hermitian operators $A$, each of which possesses a complete, orthonormal set of eigenvectors $\alpha_{i}$ and a corresponding set of real eigenvalues $a_i$ and the only possible values of any measurement of this observable is one of these eigenvalues.

Note: in the cases when the eigenvalues are discretely distributed (e. g., the eigenvalues fail to have a limit point), we get “quantized” behavior from this observable.

We’ll use observables with discrete eigenvalues unless we say otherwise.

Now: is a function of an observable itself an observable? The answer is “yes” if the function is real analytic and we assume that $(A)^n(\psi) = A(A(A....A(\psi))$. To see this: assume that $f(z) = \sum_i c_i z^i$ and note that if $A$ is an observable operator then so is $cA^n$ for all $n$. Note: one can do this by showing that the eigenvectors for $A$ do not change and that the eigenvalues merely go up by power. The completeness of the eigenvectors imply convergence when we pass to $f$.

Now we have states and observables. But how do they interact?
Remember that we showed the following:

Let $A$ be a linear operator with a complete orthonormal eigenbasis $\alpha_i$ and corresponding real eigenvalues $a_i$. Let $\psi$ be an element of the Hilbert space with unit norm and let $\psi = \sum_j b_j \alpha_j$.

Then the function $P(y = a_i) = (|b_i|)^2$ is a probability density function. (note: $b_i = \langle \alpha_i , \psi \rangle$).

This will give us exactly what we need! Basically, if the observable has operator $A$ system and is in state $\psi$, then the probability of a measurement yielding a result of $a_i$ is $(|\langle \alpha_i , \psi \rangle|)^2$ Note: it follows that if the state $\phi = \alpha_i$ then the probability of obtaining $a_i$ is exactly one.

We summarize this up by Postulate 3: (page 49 of Gillespie, stated for the “scattered eigenvalues” case):

Postulate 3: If an observable operator $A$ has eigenbasis $\alpha_i$ with eigenvalues $a_i$ and if the corresponding observable is measured on a system which, immediately prior to the measurement is in state $\psi$ then the strongest predictive statement that can be made concerning the result of this measurement is as follows: the probability that the measurement will yield $a_k$ is $(|\langle \alpha_i , \psi \rangle|)^2$.

Note: for simplicity, we are restricting ourselves to observables which have distinct eigenvalues (e. g., no two linearly independent eigenvectors have the same eigenvalues). In real life, some observables DO have different eigenvectors with the same eigenvalue (example from calculus; these are NOT Hilbert Space vectors, but if the operator is $d^2/dx^2$ then $sin(x)$ and $cos(x)$ both have eigenvalue -1. )

Where we are now: we have a probability distribution to work with which means that we can calculate an expected value and a variance. These values will be fundamental when we tackle uncertainty principles!

Just a reminder from our courses in probability theory: if $Y$ is a random variable with density function $P$

$E(Y) = \sum_i y_i P(y_i)$ and $V(Y) = E(Y^2) -(E(Y))^2$.

So with our density function $P(y = a_i) = (|b_i|)^2$ (we use $b_i = \langle \alpha_i , \psi \rangle$ to save space), then if $E(A)$ is the expected observed value of the observable (the expected value of the eigenvalues):
$E(A) = \sum_i a_i (b_i)^2$. But this quantity can be calculated in another way:

$\langle \psi , A(\psi) \rangle = \langle \sum b_i \alpha_i , A(\sum b_i \alpha_i) \rangle = \langle \sum b_i \alpha_i , \sum a_i b_i \alpha_i) \rangle = \sum_i \overline{b_i} b_i a_i \langle \alpha_i, \alpha_i \rangle = \sum_i \overline{b_i} b_i a_i = \sum_i |b_i|^2 a_i = E(A)$. Yes, I skipped some easy steps.

Using this we find $V(A) = \langle \psi, A^2(\psi) \rangle - (\langle \psi, A(\psi) \rangle )^2$ and it is customary to denote the standard deviation $\sqrt{V(A)} = \Delta(A)$

In our next installment, I give an illustrative example.

In a subsequent installment, we’ll show how a measurement of an observable affects the state and later how the distribution of the observable changes with time.

## July 11, 2011

### Quantum Mechanics for teachers of undergraduate mathematics I

I am planning on writing up a series of notes from the out of print book A Quantum Mechanics Primer by Daniel Gillespie.

My background: mathematics instructor (Ph.D. research area: geometric topology) whose last physics course (at the Naval Nuclear Power School) was almost 30 years ago; sophomore physics was 33 years ago.

Your background: you teach undergraduate mathematics for a living and haven’t had a course in quantum mechanics; those who have the time to study a book such as Quantum Mechanics and the Particles of Nature by Anthony Sudbery would be better off studying that. Those who have had a course in quantum mechanics would be bored stiff.

Topics the reader should know: probability density functions, square integrability, linear algebra, (abstract inner products (Hermitian), eigenbasis, orthonormal basis), basic analysis (convergence of a series of functions) differential equations, dirac delta distribution.

My purpose: present some opportunities to present applications to undergraduate students e. g., “the dirac delta “function” (distribution really) can be thought of as an eigenvector for this linear transformation”, or “here is an application of non-standard inner products and an abstract vector space”, or “here is a non-data application to the idea of the expected value and variance of a probability density function”, etc.

Basic mathematical objects
Our vector space will consist of functions $\psi : R \rightarrow C$ (complex valued functions of a real variable) for which $\int^{\infty}_{-\infty} \overline{\psi} \psi dx$ is finite. Note: the square root of a probability density function is a vector of this vector space. Scalars are complex numbers and the operation is the usual function addition.

Our inner product $\langle \psi , \phi \rangle = \int^{\infty}_{-\infty} \overline{\psi} \phi dx$ has the following type of symmetry: $\langle \psi , \phi \rangle= \overline{\langle \phi , \psi \rangle}$ and $\langle c\psi , \phi \rangle = \langle \psi , \overline{c} \phi \rangle = \overline{c}\langle \psi , \phi \rangle$.

Note: Our vector space will have a metric that is compatible with the inner product; such spaces are called Hilbert spaces. This means that we will allow for infinite sums of functions with some convergence; one might think of “convergence in the mean” which uses our inner product in the usual way to define the mean.

Of interest to us will be the Hermitian linear transformations $H$ where $\langle H(\psi ), \phi \rangle = \langle \psi ,H(\phi) \rangle .$ It is an easy exercise to see that such a linear transformation can only have real eigenvalues. We will also be interested in the subset (NOT a vector subspace) of vectors $\psi$ for which $||(\langle \psi , \phi \rangle)||^2 = 1$.

Eigenvalues and eigenvectors will be defined in the usual way: if $H(\psi) = \alpha \psi$ then we say that $\psi$ is an eigenvector for $H$ with associated eigenvalue $\alpha$. If there is a countable number of orthornormal eigenvectors whose “span” (allowing for infinite sums) includes every element of the vector space, then we say that $H$ has a complete orthonormal eigenbasis.

It is a good warm up exercise to show that if $H$ has a complete orthonormal eigenbasis then $H$ is Hermitian.

Hint: start with $\langle H(\psi ), \phi \rangle$ and expand $\psi$ and $\phi$ in terms of the eigenbasis; of course the linear operator $H$ has to commute with the infinite sum so there are convergence issues to be concerned about.

The outline goes something like this: suppose $\epsilon_i$ is the complete set of eigenvectors for $H$ with eigenvalues $a_i$ and $\psi = \sum_i b_i \epsilon_i$ and $\phi = \sum_i c_i \epsilon_i$
$\langle H(\psi ), \phi \rangle =\langle H(\sum_i b_i \epsilon_i ), \phi \rangle = \langle \sum_i H(b_i \epsilon_i ), \phi \rangle = \langle \sum_i b_i a_i \epsilon_i , \phi \rangle = \sum_i a_i\langle b_i \epsilon_i , \phi \rangle$

Now do the same operation on the left side of the inner product and use the fact that the basis vectors are mutually orthogonal. Note: there are convergence issues here; those that relate the switching of the infinite sum notation outside of the inner product can be handled with a dominated convergence theorem for integrals. But the intuition taken from finite vector spaces works here.

The other thing to note is that not every Hermitian operator is “closed”; that is it is possible for $\psi$ to be square integrable but for operator $H(\phi) = x \phi$ to not be square integrable.

Probability Density Functions

Let $H$ be a linear operator with a complete orthonormal eigenbasis $\epsilon_i$ and corresponding real eigenvalues $a_i$. Let $\psi$ be an element of the Hilbert space with unit norm and let $\psi = \sum_j b_j \epsilon_j$.

Claim: the function $P(y = a_i) = (|b_i|)^2$ is a probability density function. (note: $b_i = \langle \epsilon_i , \psi \rangle$).

The fact that $(|b_i|)^2 \leq 1$ follows easily from the Cauchy-Schwartz inequality. Also note that $1 = | \langle \psi, \psi \rangle | = | \langle \sum b_i \epsilon_i,\sum b_i \epsilon_i \rangle | = |\sum_i (b_i)^2 \langle \epsilon_i, \epsilon_i \rangle | = |\sum_i (b_i)^2|$

Yes, I skipped some steps that are easy to fill in. But the bottom line is that this density function now has a (sometimes) finite expected value and a (sometimes) finite variance.

With the mathematical preliminaries (mostly) out of the way, we are ready to see how this applies to physics.