Some Basic Math for Quantum Mechanics

I’ve been reading a book on quantum computing lately (Nielsen and Chuang) and I thought I’d write a hyper-condensed review of the super-fundamental bits required for pure QM, but restricted to quantum computing. I’m not sure how human-readable this is in general, but the math will be easy for anyone who’s taken an undergrad course. Of course, the disclaimer is that I mostly come from an undergrad education and lots of reading in my free time, so don’t take this as gospel.

NB: because this is in the context of quantum computing, I will make no mention of infinite-dimentional Hilbert spaces. If I were to include those, I’d take my cues from the first chapter of Shankar’s QM book instead, which is a rather weighty read. I also won’t talk about metrics because that’s general relativity land, and there be dragons.

Lastly, this won’t be nearly the full material - more of a reminder for myself of the results and some background stuff for convenience.

Notation

$A^\top$ is the transpose of a column vector or matrix $A$.

$A^*$ is the complex conjugate of the elements of a vector or matrix $A$.

$A^\dagger = A^{* ^{\top }}$ (AKA “dagger”) is the conjugate transpose of a vector or matrix $A$. Lots of literature uses the asterisk to represent this, which cost me some homework points back when I was learning this and didn’t know any better. The asterisk will now and forever mean complex conjugate.

Vector Spaces

We denote a vector as $\lvert v\rangle$, or $v$ for short if it’s obvious what we’re talking about. Say $V$ is a set of vectors. Then $V$ is a vector space if for any $\lvert u\rangle, \lvert v\rangle, \lvert w\rangle \in V$ and $a, b \in \mathbb{C}$:

1. Addition is associative: $u + (v + w) = (u + v) + w$
2. Addition is commutative: $u + v = v + u$
3. There exists $0 \in V$ s.t. $v + 0 = v$. (Note that the $0$ here is shorthand for the vector $\lvert 0\rangle$, not a scalar $0$.)
4. Additive inverse: there exists $-v \in V$ s.t. $v + (-v) = 0$.
5. Closed under linear combinations: $a u + b v \in V$.
6. …and a couple of others.

We normally think of a vector $\lvert v\rangle$ in an $n$-dimensional vector space $V$ as an n-tuple of numbers, i.e., the column vector $(v_1, ..., v_n)^\top$, where $\lvert v\rangle = \sum_i v_i \lvert i\rangle$ for a particular orthonormal basis $\lvert i\rangle$. Change basis and the tuple representing the vector in that basis will be different.

A vector can also be a function or even a matrix, as long as it satisfies the standard properties, but, for our purposes, we’ll use the tuple representation.

Bases

A set of vectors $\lvert v_1\rangle, \ldots, \lvert v_n\rangle$ is linearly independent if no vector in this set can be written as a linear combination of the others, i.e., for any coefficients $a_i$,

$$\sum_i a_i \lvert v_i\rangle = 0 \implies a_i = 0$$

If a set of $n$ linearly independent vectors spans the given vector space $V$ (i.e., every vector $\lvert v\rangle$ in $V$ can be written as a linear combination of these vectors), then it’s a basis. The number of vectors in the basis is the dimension of the space.

Linear Operators

An operator $A$ is a map $A : V \rightarrow W$, i.e., it takes a vector in $V$ and maps it to a vector in $W$. All our operators will be linear:

$$A \sum_i v_i \lvert i\rangle = \sum_i v_i A \lvert i\rangle$$

So operators distribute into and out of sums and everything is peachy. The identity map $I$ is customarily written without reference to any particular basis or dimensionality and is understood to take any vector and scalar into itself.

Importantly, if we have a basis $\lvert v_i\rangle$ of $V$ and basis $\lvert w_i\rangle$ of W, we recognize that A operating on some vector $\lvert v_j\rangle$ produces some linear combination of $W$’s basis vectors, which we call $A$’s matrix elements. This indicates that

1. the operator has a representation as a matrix, and
2. this representation is basis-dependent.

Inner Products and Dual Vectors

The inner products of two vectors $\lvert w\rangle$ and $\lvert v\rangle$, respectively, is written as $\langle w\lvert v\rangle$. This is Dirac’s bra-ket notation and, mysteriously, is never used outside physics. Specifically, the $\lvert v\rangle$ is a “ket” and the $\langle w\lvert$ is a “bra”.

The inner product of two vectors $\lvert u\rangle$ and $\lvert v\rangle$, respectively, is defined (assuming we represent our vectors as n-tuples) as

$$\langle u\lvert v\rangle = \lvert u\rangle^\dagger \lvert v\rangle = \sum_i u_i^* v_i$$

where the two vectors are represented in the same orthonormal basis and $u_i$ and $v_i$ are their coefficients.

The magnitude of a vector $\lvert v\rangle$ is written as

$$\lvert \lvert v \rvert \rvert = \sqrt{\langle v\lvert v\rangle}$$

This thing is obviously real and non-negative if we believe our definition of the inner product.

($\langle w\lvert$, strictly speaking, is a dual vector to $\lvert w\rangle$. It’s a map $\langle w\lvert : V \rightarrow \mathbb{C}$ that takes a vector and turns it into a number. If you have an orthonormal basis of vectors $\lvert v_i\rangle$ for vector space $V$, we can define a dual basis to it as the set $\langle w_i\rvert$ such that $\langle w_i\lvert v_j\rangle = \delta^i_j$. Again, not delving into metrics here. That discussion can be found in innumerable sources, e.g., Wald’s general relativity book.)

Gram-Schmidt Orthonormalization

Leaving this here for future reference: given a potentially non-orthonormal basis $\lvert v\rangle$, we can turn it into an orthonormal one $\lvert w\rangle$. First, take $\lvert w_1\rangle = \frac{\lvert v_1\rangle}{\lvert\lvert v_1\rvert\rvert}$. Then, for $i = 2 \ldots n$, take the original $\lvert v_i\rangle$, subtract its projection onto all the $\lvert w_j\rangle$s we’ve calculated, then normalize. This iterative procedure gives us $\lvert w_i\rangle$.

So for $i = 2 \ldots n$:

$$\lvert w_i\rangle = \frac{\lvert v_i\rangle - \sum_{j=1}^{i-1} \langle w_j \lvert v_i\rangle \lvert w_j\rangle}{\lvert\lvert \cdot \rvert\rvert}$$

where the denominator is just the magnitude of the numerator, for the sake of clean-ish notation.

Outer Products, Projection Operators, Completeness Relation

An outer product of two vectors $\lvert w\rangle \in W$ and $\lvert v\rangle \in V$ is written as $\lvert w\rangle\langle v\rvert$. This is a linear operator; it’s also a map $V \rightarrow W$.

Suppose we have vector $\lvert v\rangle = \sum_i v_i \lvert i\rangle$. Let’s operate on it with the quantity $\sum_i \lvert i\rangle\langle i\rvert$:

$$\Big(\sum_i \lvert i\rangle\langle i\rvert\Big) \lvert v\rangle = \sum_i \lvert i\rangle\langle i\rvert v\rangle = \sum_i v_i\lvert i\rangle = \lvert v\rangle$$

It must be, then, that $\sum_i \lvert i\rangle\langle i\rvert = I$. So we can dump it in the middle of any string of vector-operator operations without changing the result.

For instance, let’s take operator $A : V \rightarrow W$ and find its representation with respect to a particular basis $\lvert v_i\rangle$ of $V$ and $\lvert w_j\rangle$ of $W$:

$$A = I A I = \Big(\sum_i \lvert v_i\rangle\langle v_i\lvert \Big) A \Big(\sum_j \lvert w_j\rangle\langle w_j\rvert \Big) = \sum_{ij} \langle v_i \lvert A \rvert w_j \rangle \lvert v_i\rangle\langle w_j\rvert$$

Here $\langle v_i \lvert A \rvert w_j \rangle$ is the matrix element of $A$. This is a scalar.

A single term P = $\lvert i\rangle\langle i\rvert$ is a projection operator onto the $i$th basis element; P = $\sum_{j \in S} \lvert j\rangle\langle j\rvert = I$ for some set of indices $S$ is a projection onto that subspace. One can verify that $P P = P$.

(More later.)