Fundamental Mathematics of Quantum Computing I

Fundamental Mathematics of Quantum Computing I

Quantum computing is often introduced through circuits: Hadamard gates, CNOTs, measurements, and perhaps a few algorithms.

That is useful, but it hides the structure underneath.

A quantum state is not fundamentally a bit string. It is a vector in a complex Hilbert space. A measurement is not fundamentally a readout operation. It is tied to the spectral structure of a Hermitian operator. A quantum gate is not merely a box in a circuit diagram. It is a unitary transformation preserving the geometry of the state space.

And once global phase is removed, even the space of states itself becomes geometric.

1. Quantum states live in Hilbert space

Let $\mathcal H$ be a complex Hilbert space.

A pure quantum state is represented by a normalized vector

$$|\psi\rangle \in \mathcal H,$$

satisfying

$$\langle \psi|\psi\rangle = 1.$$

For a single qubit,

$$\mathcal H = \mathbb C^2,$$

with computational basis

$$|0\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix}, \qquad |1\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix}.$$

A general qubit is therefore

$$|\psi\rangle = \alpha |0\rangle + \beta |1\rangle,$$

where

$$|\alpha|^2+|\beta|^2=1.$$

The coefficients are amplitudes, not probabilities.

Probabilities appear only after taking absolute squares.

2. Inner products determine the geometry

The inner product between two states is written

$$\langle \phi|\psi\rangle.$$

It determines both lengths and angles.

The norm of a vector is

$$\|\psi\| = \sqrt{\langle\psi|\psi\rangle}.$$

Two states are orthogonal when

$$\langle\phi|\psi\rangle=0.$$

For example,

$$\langle0|1\rangle=0.$$

The computational basis therefore consists of two orthogonal directions in $\mathbb C^2$.

More generally, if $\{|i\rangle\}$ is an orthonormal basis,

$$\sum_i |i\rangle\langle i| = I.$$

Hence every state may be reconstructed as

$$|\psi\rangle = \sum_i |i\rangle\langle i|\psi\rangle.$$

The coefficient of $|i\rangle$ is simply

$$\langle i|\psi\rangle.$$

3. Linear operators act on quantum states

A linear operator

$$A:\mathcal H\rightarrow\mathcal H$$

satisfies

$$A\left( \alpha|\psi\rangle+\beta|\phi\rangle \right) = \alpha A|\psi\rangle + \beta A|\phi\rangle.$$

The adjoint of $A$ is denoted by

$$A^\dagger.$$

For a finite-dimensional matrix,

$$A^\dagger = \overline A^{\,T}.$$

For example,

$$A= \begin{pmatrix} 1&i\\ 2&3 \end{pmatrix}$$

has adjoint

$$A^\dagger = \begin{pmatrix} 1&2\\ -i&3 \end{pmatrix}.$$

This operation separates two important classes of quantum operators.

4. Hermitian operators describe observables

An operator is Hermitian when

$$A=A^\dagger.$$

The significance of this condition is spectral.

Suppose

$$A|\psi\rangle = \lambda|\psi\rangle.$$

Then

$$\langle\psi|A|\psi\rangle = \lambda\langle\psi|\psi\rangle.$$

Since $A$ is Hermitian,

$$\langle\psi|A|\psi\rangle = \langle A\psi|\psi\rangle.$$

Using

$$A|\psi\rangle = \lambda|\psi\rangle,$$

taking the adjoint gives (This is because scalars are complex conjugated when moved to the bra)

$$\langle A\psi|=\lambda^*\langle\psi|$$

we obtain

$$\langle A\psi|\psi\rangle = \lambda^* \langle\psi|\psi\rangle.$$

Therefore,

$$\lambda \langle\psi|\psi\rangle = \lambda^* \langle\psi|\psi\rangle.$$

Since $|\psi\rangle\neq0$,

$$\lambda=\lambda^*.$$

Hence

$$\lambda\in\mathbb R$$

The eigenvalues of a Hermitian operator are real.

This is precisely what we need if eigenvalues are to represent possible measurement outcomes.

5. Distinct eigenvalues produce orthogonal states

Let

$$A|\psi\rangle = \lambda|\psi\rangle$$

and

$$A|\phi\rangle = \mu|\phi\rangle,$$

where

$$\lambda\neq\mu.$$

Then

$$\langle\phi|A|\psi\rangle = \lambda \langle\phi|\psi\rangle.$$

Hermiticity also gives

$$\langle\phi|A|\psi\rangle = \langle A\phi|\psi\rangle = \mu \langle\phi|\psi\rangle.$$

Thus

$$(\lambda-\mu) \langle\phi|\psi\rangle = 0.$$

Because $\lambda\neq\mu$,

$$\langle\phi|\psi\rangle=0$$

Eigenstates corresponding to distinct eigenvalues are orthogonal.

This is more than a convenient algebraic fact. It means that different measurement outcomes correspond to geometrically distinguishable directions in Hilbert space.

6. Measurement and the Born rule

Suppose

$$A|a_n\rangle = a_n|a_n\rangle$$

for an observable $A$.

If the system is in state $|\psi\rangle$, the probability of obtaining the measurement outcome $a_n$ is

$$P(a_n) = |\langle a_n|\psi\rangle|^2.$$

For a qubit

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,$$

measurement in the computational basis gives

$$P(0)=|\alpha|^2$$

and

$$P(1)=|\beta|^2.$$

The condition

$$|\alpha|^2+|\beta|^2=1$$

is therefore exactly the normalization of total probability.

7. Unitary operators preserve quantum states

A unitary operator satisfies

$$U^\dagger U = UU^\dagger = I.$$

Equivalently,

$$U^{-1}=U^\dagger.$$

Let

$$|\psi'\rangle = U|\psi\rangle.$$

Then

$$\langle\psi'|\psi'\rangle = \langle\psi| U^\dagger U |\psi\rangle.$$

Since

$$U^\dagger U=I,$$

we have

$$\langle\psi'|\psi'\rangle = \langle\psi|\psi\rangle.$$

Thus

$$\boxed{ \|U\psi\|=\|\psi\| }.$$

Unitary transformations preserve the normalization of quantum states.

Geometrically, they preserve inner products, lengths, and angles.

8. Commutators and compatible structure

For two operators $A$ and $B$, define the commutator

$$[A,B] = AB-BA.$$

If

$$[A,B]=0,$$

then the operators commute.

For commuting Hermitian operators, one may choose a common eigenbasis.

To see the basic idea, suppose

$$A|a\rangle = a|a\rangle.$$

Then

$$A(B|a\rangle) = AB|a\rangle.$$

If $AB=BA$,

$$A(B|a\rangle) = BA|a\rangle = aB|a\rangle.$$

So $B|a\rangle$ remains inside the eigenspace of $A$ associated with eigenvalue $a$.

This is the algebraic reason simultaneous diagonalization becomes possible.

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9. Composite quantum systems require tensor products

A single qubit lives in

$$\mathbb C^2.$$

Two qubits do not live in another copy of $\mathbb C^2$.

Their joint state space is

$$\mathbb C^2\otimes\mathbb C^2.$$

In general,

$$\dim(V\otimes W) = \dim(V)\dim(W).$$

Therefore,

$$\dim( \mathbb C^2\otimes\mathbb C^2 ) = 4.$$

The natural basis is

$$|00\rangle, \quad |01\rangle, \quad |10\rangle, \quad |11\rangle.$$

Explicitly,

$$|00\rangle = \begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix}, \qquad |01\rangle = \begin{pmatrix} 0\\ 1\\ 0\\ 0 \end{pmatrix},$$ $$|10\rangle = \begin{pmatrix} 0\\ 0\\ 1\\ 0 \end{pmatrix}, \qquad |11\rangle = \begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix}.$$

A general two-qubit state is

$$|\psi\rangle = c_{00}|00\rangle + c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle,$$

with

$$|c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1.$$

For $n$ qubits,

$$\mathcal H_n = (\mathbb C^2)^{\otimes n},$$

and therefore

$$\dim\mathcal H_n=2^n$$

10. Operators also tensor together

If $A$ acts on one subsystem and $B$ acts on another, the combined operator is

$$A\otimes B.$$

For

$$A= \begin{pmatrix} a&b\\ c&d \end{pmatrix},$$

we have

$$A\otimes B = \begin{pmatrix} aB&bB\\ cB&dB \end{pmatrix}.$$

This is the Kronecker product representation of the tensor product.

It provides the basic language for writing operations on multi-qubit systems.

11. The Pauli operators

Three matrices appear repeatedly in the structure of a qubit:

$$X= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix},$$ $$Y= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix},$$

and

$$Z= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}.$$

Each is Hermitian and unitary.

Their eigenvalues are

$$\pm1.$$

They also satisfy the commutation relations

$$[\sigma_i,\sigma_j] = 2i\epsilon_{ijk}\sigma_k.$$

In particular,

$$[X,Y]=2iZ,$$ $$[Y,Z]=2iX,$$

and

$$[Z,X]=2iY.$$

The Pauli matrices therefore encode both observables and transformations within the same algebraic structure.

12. Global phase is physically redundant

Consider

$$|\psi\rangle$$

and

$$e^{i\alpha}|\psi\rangle.$$

For any state $|\phi\rangle$,

$$\left| \langle\phi| e^{i\alpha} |\psi\rangle \right|^2 = |e^{i\alpha}|^2 |\langle\phi|\psi\rangle|^2.$$

Since

$$|e^{i\alpha}|=1,$$

we obtain

$$\left| \langle\phi| e^{i\alpha} |\psi\rangle \right|^2 = |\langle\phi|\psi\rangle|^2.$$

Thus

$$|\psi\rangle \sim e^{i\alpha}|\psi\rangle$$

Vectors that differ only by a global phase represent the same physical state.

So a physical state is not quite a vector.

It is a ray.

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13. Quantum states form a projective space

Let

$$z\sim\lambda z, \qquad \lambda\in\mathbb C^\*.$$

Then the space of rays in $\mathbb C^m$ is

$$\boxed{ \mathbb{CP}^{m-1} = \frac{ \mathbb C^m\setminus\{0\} }{ \mathbb C^\* } }.$$

A point in complex projective space can be written using homogeneous coordinates

$$[z_0:z_1:\cdots:z_n].$$

The equivalence relation is

$$[z_0:\cdots:z_n] = [\lambda z_0:\cdots:\lambda z_n]$$

for every nonzero

$$\lambda\in\mathbb C.$$

This quotient is the natural state space once physically irrelevant scaling is removed.

14. The geometry of a single qubit

For a qubit,

$$|\psi\rangle = z|0\rangle+w|1\rangle,$$

with

$$|z|^2+|w|^2=1.$$

The pair

$$(z,w)\in\mathbb C^2\simeq\mathbb R^4$$

therefore lies on

$$S^3.$$

But states differing by global phase are equivalent:

$$(z,w) \sim (e^{i\alpha}z,e^{i\alpha}w).$$

Write

$$z=r_0e^{i\phi_0}, \qquad w=r_1e^{i\phi_1}.$$

Multiplying the whole state by

$$e^{-i\phi_0}$$

gives

$$|\psi\rangle \sim r_0|0\rangle + r_1e^{i(\phi_1-\phi_0)}|1\rangle.$$

Only the relative phase remains.

Let

$$r_0=\cos\theta, \qquad r_1=\sin\theta.$$

Then

$$\boxed{ |\psi\rangle = \cos\theta|0\rangle + e^{i\phi}\sin\theta|1\rangle }.$$

The pure-state space of a qubit is therefore

$$\mathbb{CP}^1.$$

And geometrically,

$$\mathbb{CP}^1 \simeq S^2$$

This is the Bloch sphere.

15. From $U(2)$ to $SU(2)$

The unitary group is

$$U(2) = \{ U\in M_2(\mathbb C) : U^\dagger U=I \}.$$

Restricting to unit determinant gives the special unitary group

$$SU(2) = \{ U\in U(2): \det U=1 \}.$$

Every matrix in $SU(2)$ can be written as

$$U= \begin{pmatrix} z&w\\ -\overline w&\overline z \end{pmatrix},$$

with

$$|z|^2+|w|^2=1.$$

Write

$$z=a+ib, \qquad w=c+id.$$

Then

$$a^2+b^2+c^2+d^2=1.$$

Thus

$$\boxed{ SU(2)\simeq S^3 }.$$

The same three-sphere that appeared when normalizing a two-component complex state now appears as a Lie group of transformations.

16. Quaternions and $SU(2)$

A quaternion has the form

$$q=a+bi+cj+dk,$$

where

$$i^2=j^2=k^2=-1.$$

The multiplication rules include

$$ij=k, \qquad jk=i, \qquad ki=j,$$

while reversing the order changes the sign.

The conjugate is

$$\bar q = a-bi-cj-dk.$$

Its norm satisfies

$$q\bar q = a^2+b^2+c^2+d^2.$$

Unit quaternions therefore satisfy

$$a^2+b^2+c^2+d^2=1.$$

Hence the unit quaternions also form

$$S^3.$$

This gives another realization of

$$SU(2).$$

17. From $SU(2)$ to rotations

Consider a purely imaginary quaternion

$$v=xi+yj+zk.$$

Such quaternions may be identified with

$$\mathbb R^3.$$

For a unit quaternion $q$, define

$$R_q(v) = qv\bar q.$$

Because

$$q^{-1}=\bar q$$

for a unit quaternion,

$$R_q(v) = qvq^{-1}.$$

This transformation preserves lengths and acts as a rotation of $\mathbb R^3$.

It gives a map

$$SU(2)\longrightarrow SO(3).$$

But

$$q$$

and

$$-q$$

produce the same rotation:

$$R_q=R_{-q}.$$

Therefore the map is two-to-one.

This is the familiar double-cover relation

$$SU(2)\rightarrow SO(3)$$

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18. The Hopf fibration

Several spaces that have appeared separately now fit together:

$$S^1\simeq U(1),$$ $$S^3\simeq SU(2),$$

and

$$S^2\simeq\mathbb{CP}^1.$$

They are related by the Hopf fibration

$$S^1 \longrightarrow S^3 \longrightarrow S^2$$

For a normalized qubit, $S^3$ contains the state vector together with its global phase.

Quotienting by the phase

$$S^1\simeq U(1)$$

leaves the physical state space

$$S^2.$$

Schematically,

$$\frac{S^3}{S^1} \simeq S^2.$$

The Bloch sphere is therefore not merely a convenient visualization.

It arises naturally from removing the physically redundant $U(1)$ phase from the normalized state space.

19. Why Lie groups appear naturally

A group $G$ has

• an associative multiplication,
• an identity,
• and inverses.

A Lie group is additionally a smooth manifold for which multiplication and inversion are smooth.

Thus

$$\text{Lie group} = \text{group} + \text{smooth manifold}$$

Groups such as

$$U(n), \qquad SU(n), \qquad SO(n)$$

are therefore simultaneously algebraic objects and geometric spaces.

Their local structure near the identity is captured by the tangent space

$$T_eG.$$

This tangent space is the Lie algebra of the group:

$$\boxed{ \mathfrak g = T_eG }.$$

This point of view turns continuous quantum transformations into geometry.

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20. The emerging picture

The mathematics developed above can be organized into three layers.

State space

$$|\psi\rangle\in\mathcal H$$

with

$$\langle\psi|\psi\rangle=1.$$

After removing global phase,

$$\text{physical states} = \mathbb P(\mathcal H).$$

For a qubit,

$$\mathbb P(\mathbb C^2) = \mathbb{CP}^1 \simeq S^2.$$

Observables

$$A=A^\dagger.$$

Hermitian operators have real eigenvalues and orthogonal eigenspaces.

Measurement probabilities are determined by projection amplitudes.

Transformations

$$U^\dagger U=I.$$

Unitary operators preserve the geometry of Hilbert space.

For a qubit, the structure naturally leads to

$$SU(2)\simeq S^3,$$

with

$$SU(2)\longrightarrow SO(3)$$

connecting quantum transformations to ordinary rotations.