Quantum State Spaces

These are some old, and pretty raw, notes I took while studying quantum mechanics.

Mathematical Spaces

Vector Space

As a reminder, the definition of a vector space is as follows. A vector space is a set $V$ and a field $\mathbb F$ (e.g. $\mathbb R$ or $\mathbb C$) with two operations

1. Vector addition $+:V\times V \mapsto V$
2. Scalar multiplication $\cdot:\mathbb F\times V \mapsto V$

that satisfy the following properties

Inner-product Space

An inner-product space is a vector space equipped with a binary function $\inp \cdot \cdot: V\times V \mapsto \mathbb F$, called the inner-product that satisfies the following properties

For the last property conjugate symmetry ensures that the result is real since $\inp {\mathbf x} {\mathbf x} = \inp{\mathbf x}{\mathbf x}^*$, so the result must be real. These properties also imply that \(\inp{\mathbf 0}{\mathbf 0} = 0\). This follows from the linearity of the inner product.

It is also easy to show that the following properties hold

• $\langle \mathbf x, \mathbf y + \mathbf z\rangle = \langle \mathbf x, \mathbf y \rangle + \langle\mathbf x, \mathbf z\rangle$
• $\langle \mathbf x, a \mathbf y \rangle = a^*\langle \mathbf x, \mathbf y \rangle$

Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality states that

Taking square roots of both sides (and preempting the next section slightly) we can see that this is effectively saying that the magnitude of the inner-product of two vectors, must be less than the geometric mean of the two magnitudes. Intuitively, think of the scalar product of ordinary $\mathbb R^n$ vectors, if we increase the length of $\mathbf x$ the dot product gets larger, but it is always scaled down by $\mathbf y$ and vice-versa.

This lets us extend the notion of angles between vectors to abstract inner-product spaces with real inner products as

where the Cauchy-Schwarz inequality ensures the right-hand side of this equation is in $[-1, 1]$.

Normed Vector Space

The square-root of the inner-product of a vector with itself defines a norm

and thus we also have a normed vector space.

The norm must obey the following properties

Properties 1. & 2. follow simply from the properties of the inner-product and 3. can be proven using the Cauchy-Schwarz ineq.

Metric Space

Since we have a norm we can also define a metric $\delta \in \mathbb R \geq 0$

that satisfies all the required properties of a metric

Property 3. follows directly from the definition of $\delta$ and the fact the norm obeys the triangle inequality.

Proof of 1:

The direction $\mathbf x = \mathbf y \Rightarrow \delta(\mathbf x, \mathbf y) = 0$ is clear;

In the other direction

Proof of 2:

Use the properties of the inner product to pull a minus sign out of each argument

where we can also ignore the squares since $\delta(\mathbf x, \mathbf y) \geq 0$.

Interestingly non-negativity can be proved from the three properties above

which implies $\delta(\mathbf x, \mathbf y) \geq 0$.

Cauchy Sequences & Completeness

In a metric space $M$ a sequence $x_1, x_2, ...$ is ‘Cauchy’ if for every $\epsilon \in \mathbb R$ there exists $m, n \in \mathbb N$ such that $\delta(x_m, x_n) < \epsilon$. Roughly, the sequence is getting closer and closer. Note however that the limit of this sequence is not always in $M$. If the limits of all Cauchy sequences of elements of $M$ themselves lie in $M$, then $M$ is said to be complete.

Hilbert Spaces

A Hilbert space is a complete inner-product space. We can see that being an inner-product space is already quite a strong statement since it implies the space is also a normed vector space and a metric space. Despite being completely abstract we can also discuss things such as the angle between vectors, their lengths, and distances between them.

Some usual examples of Hibert spaces are

Euclidean space: The vectors are just ordinary vectors in $\mathbb R^N$, the inner product is the scalar product.

Continuous functions: The vectors are the functions with a suitable basis (e.g. Fourier decomposition), the inner product can be defined as an integral

Dual Space & Bra-Ket Notation

A linear function $\phi : V \mapsto \mathbb F$, has the property

The set of all such linear functions itself forms a vector space where the vectors are the functions themselves;

• Scalar multiplication: $a\phi(\cdot)$
• Vector addition: $\phi(\cdot) + \rho(\cdot)$

It is clear that all the required properties for a vector space are fulfilled by linear functions. We can also identify every linear function $V \mapsto \mathbb F$ with a vector in $V$. Thus the space of linear functions forms a so-called “dual” vector space to $V$, which is denoted by $V^*$.

For something like $\mathbb R^N$ Euclidean space the identification is clear since every linear function is just a linear combination of the components of $\mathbf v$:

This is just the dot product of $\mathbf v$ with some vector $\boldsymbol \phi$, with components $\phi_i$. The situation is very similar for functions; now we can just use the inner product defined above (an integral is kinda like a big sum). For more abstract vector spaces it’s not so clear to me how we can know that each linear function is identified with a vector, it appears quite technical.

The inner-product is a linear function in the first argument, thus we can identify $\langle\cdot,\mathbf y\rangle$ with some element of $V^*$, this is the element associated with vector $\mathbf y$. In Dirac’s bra-ket notation the vector $\mathbf x$ is denoted $|\mathbf x\rangle\in V$ (a ‘ket’) and the corresponding dual vector is denoted $\langle \mathbf x|\in V^*$ (a ‘bra’). Then the application of the linear function on a vector is represented by juxtaposition, but we elide the extra $|$

Also, if we have some operator that acts on the states then

Classical vs. Quantum State-spaces

In classical physics we often describe the phase-space or state-space. What is different from the quantum mechanical view in this case?

Example: Two States

To answer this its easiest to step back somewhat and consider the simplest possible system; a classical system that can only be in two states, labelled $a$ and $b$. In this case, our state space is simply $\Omega = \{a, b\}$, it isn’t even 1-dimensional, it’s just a finite set. But since the system is classical we know it is definitely in one and only one of the two states. To fully describe the state of the system we need one piece of information; “$a$ or $b$?”.

Now consider the quantum analog; a system that can be measured to be in either of two states, also labelled $a$ and $b$. In this case we have a probability to measure $a$, and a probability to measure $b$. This is clearly different from the classical case where the system is in some definite state. The states $a$ and $b$ are actually the (orthogonal) axes of our Hilbert state-space, so our state space is 2-dimensional. The position in $\mathcal H$ tells us the relative probabilities of observing state $a$ or state $b$.

Compare the classical and quantum cases; in both we have a system that can be measured to be in one of two states. However, in the classical our state-space is a single binary bit, whereas in the quantum case it is a 2-dimensional (complex) space.

Example: Position in 1-dimension

Consider a particle that is located at some point in one dimension. Classically we need just one real number to describe the whole state; namely the position of the particle. For a quantum system we have a probability to measure the particle at every position, and this probability is set by some (complex) number, i.e. the state is specified by a complex function on the real line. Therefore every position has an associated complex number, perhaps somewhat counter-intuitively, each position we can measure is an orthogonal axis in our Hilbert space, meaning in this case our quantum state-space is uncountably infinite dimensional!

Comparing the quantum and the classical cases here, we go from one (real) number describing the whole system classically, to uncountably infinitely many (complex) numbers describing the whole quantum system.

These two examples hopefully make clear quite how much bigger quantum state-spaces are than classical. But they mainly serve to illustrate the fact that the possible states we can measure a system to be in are orthogonal, it is merely that in the classical case this does not matter since only one state ever has non-zero probability to be observed (i.e. probability 1) and thus we can just identify the labels of the states with the state-space. This is not the case in quantum mechanics.

Composite Systems

Classical Systems $\Rightarrow$ Direct Sum

In classical mechanics if we compose two systems $A$ and $B$ the possible states the combined system can be in is given by the cartesian product of the component state spaces $A \times B$. Thus the number of possible states is $|A||B|$. However, to fully describe which state we are in we need only $\mathrm{dim}(A) + \mathrm{dim}(B)$ numbers to describe which state we are in, the state-space is the direct sum $A\oplus B$, where $\mathrm{dim}(A\oplus B) = \mathrm{dim}(A) + \mathrm{dim}(B)$.

Quantum Systems $\Rightarrow$ Tensor Product

In quantum mechanics the compositions of systems is given by the tensor product of their state spaces. If system A has measurable states $A$ and system B has measurable states $B$ then the state space for each isolated system is $\mathcal{H}_A$ and $\mathcal{H}_B$ with dimensions $\mathrm{dim}(\mathcal{H}_A) = |A|$ and $\mathrm{dim}(\mathcal{H}_B)=|B|$. As for the classical system the combined system has possible observable states in $A \times B$, but until we observe we do not know exactly which state it is in, so we must track the probabilities of being in each of the $|A||B|$ possible states. Thus the state space is the tensor product of the Hilbert spaces $\mathcal{H}_A \otimes \mathcal{H}_B$, which has $\mathrm{dim}(\mathcal{H}_A \otimes \mathcal{H}_B) = \mathrm{dim}(\mathcal{H}_A)\,\mathrm{dim}(\mathcal{H}_B)$.

Example

Let $A = \{a_1, a_2, a_3\}$, $B=\{b_1, b_2, b_3\}$. The possible states for a classical system to be in are

It has nine possible values, $|A||B|=3\times 3 = 9$, but we need only two trinary digits to specify which state we’re in, $\mathrm{dim}(A) + \mathrm{dim}(B) = 1 + 1 = 2$.

For the analogous quantum system the possible measurable states are also nine, but we need to track the coefficients of all of them to know their relative probabilities

with $\gamma_{ij} \in \mathbb{C}$