Complex Vector Spaces

A complex vector space is a vector space whose scalars are complex numbers. This is the default setting of nonrelativistic quantum mechanics: state vectors, amplitudes, inner products, and unitary time evolution all use complex scalar multiplication.

The phrase “complex vector space” is not just a decoration on real linear algebra. Complex linearity, conjugate linearity, and phase are distinct structures, and confusing them is a reliable way to lose signs, phases, or adjoints.

Definition

A vector space $V$ over $\mathbb C$ has vector addition and scalar multiplication by complex numbers. If $v,w\in V$ and $a,b\in\mathbb C$, then

$$av+bw\in V.$$

The scalars obey the usual complex-number rules, including multiplication by $i$ with $i^2=-1$. A complex linear combination is therefore allowed to carry both magnitude and phase.

For example, $\mathbb C^n$ is a complex vector space with vectors

$$v = \begin{pmatrix} v_1\\ \vdots\\ v_n \end{pmatrix}, \qquad v_j\in\mathbb C.$$

The complex dimension of $\mathbb C^n$ is $n$.

Complex vs Real Linearity

Every complex vector space can also be viewed as a real vector space by restricting scalars from $\mathbb C$ to $\mathbb R$. As a real vector space, $\mathbb C^n$ has dimension $2n$, because each complex coordinate has a real and imaginary part.

The converse is not automatic. A real vector space becomes a complex vector space only after one specifies how multiplication by $i$ acts. This operation is a real-linear map $J$ satisfying

$$J^2=-I.$$

Complex scalar multiplication is then built from

$$(a+ib)v = av+bJv, \qquad a,b\in\mathbb R.$$

This distinction matters because a map may be real-linear without being complex-linear. Complex conjugation on $\mathbb C$ is the simplest example:

$$K(z)=z^*.$$

It satisfies real linearity,

$$K(ax+by)=aK(x)+bK(y), \qquad a,b\in\mathbb R,$$

but not complex linearity, since

$$K(iz) = (iz)^* = -iz^* \ne iK(z)$$

for nonzero $z$.

Complex-Linear Maps

A map $T:V\to W$ between complex vector spaces is complex-linear when

$$T(av+bw) = aT(v)+bT(w)$$

for all complex scalars $a,b$. This is the linearity used for ordinary quantum operators. If an operator is complex-linear, its action on a superposition is determined by its action on the components, including their phases.

Complex conjugation is instead conjugate-linear, or antilinear:

$$K(av+bw) = a^*K(v)+b^*K(w).$$

Antilinear maps are not mistakes; they appear in time-reversal and Wigner symmetry theory. They are different objects from ordinary complex-linear operators and should not be silently mixed with them. See Antiunitary Symmetries, First Look for the operator mathematics and Antiunitary Symmetries for the physics context.

Phase and Interference

Complex scalar multiplication encodes phase. Multiplying a vector by a phase $e^{i\alpha}$ changes the vector representative but, for a pure quantum state, not the physical ray:

$$\lvert\psi\rangle \sim e^{i\alpha}\lvert\psi\rangle.$$

Relative phases are different. In a two-dimensional Hilbert space, the states

$$\lvert+\rangle = \frac{1}{\sqrt2} \left( \lvert0\rangle+\lvert1\rangle \right), \qquad \lvert-\rangle = \frac{1}{\sqrt2} \left( \lvert0\rangle-\lvert1\rangle \right)$$

are not the same ray. They give the same probabilities in the $\{\lvert0\rangle,\lvert1\rangle\}$ basis, but different probabilities in the $\{\lvert+\rangle,\lvert-\rangle\}$ basis. This is why relative phase belongs to the physical content of a superposition; see Superposition and Relative Phase.

Duals and Bras

The algebraic dual $V^*$ of a complex vector space consists of complex-linear maps

$$\ell:V\to\mathbb C.$$

Thus

$$\ell(av+bw) = a\ell(v)+b\ell(w).$$

In a complex inner-product space, the inner product connects vectors and dual vectors. With the physics convention,

$$\langle\phi\vert\psi\rangle$$

is conjugate-linear in $\phi$ and linear in $\psi$. Therefore the map

$$\lvert\phi\rangle \mapsto \langle\phi\vert$$

is conjugate-linear. For example,

$$a\lvert\phi\rangle \mapsto a^*\langle\phi\vert.$$

This is one of the main reasons that “turn a column into a row” is not enough. One must also conjugate coefficients according to the chosen convention.

Finite-Dimensional Quantum Example

A finite $n$-level quantum system uses a complex Hilbert space isomorphic to $\mathbb C^n$. In an orthonormal basis,

$$\lvert\psi\rangle = \sum_{j=1}^n c_j\lvert j\rangle, \qquad c_j\in\mathbb C.$$

Normalization is

$$\sum_{j=1}^n \lvert c_j\rvert^2=1.$$

The phases of the coefficients matter through interference. Hermitian observables and unitary evolution depend on complex conjugation in their adjoint operation, which is why complex Hilbert spaces naturally support the usual spectral and probability rules.

Real and quaternionic variants of quantum theory can be studied mathematically, but the standard formalism used throughout these pages is complex Hilbert-space quantum mechanics.

Worked Example

Define

$$T:\mathbb C^2\to\mathbb C^2, \qquad T \begin{pmatrix} z_1\\ z_2 \end{pmatrix} = \begin{pmatrix} z_1+z_2\\ iz_2 \end{pmatrix}.$$

Then $T$ is complex-linear. For $a,b\in\mathbb C$ and $u,v\in\mathbb C^2$,

$$T(au+bv) = aT(u)+bT(v),$$

because each component uses only addition and multiplication by fixed complex constants.

By contrast, the map

$$C \begin{pmatrix} z_1\\ z_2 \end{pmatrix} = \begin{pmatrix} z_1^*\\ z_2^* \end{pmatrix}$$

is not complex-linear. It is conjugate-linear:

$$C(au+bv) = a^*C(u)+b^*C(v).$$

Both maps are useful, but only $T$ is an ordinary complex-linear operator.

Common Mistakes

• Treating a complex vector space as merely a real vector space with twice as many coordinates.
• Forgetting that complex linearity must hold for all complex scalars, not just real scalars.
• Calling complex conjugation linear without specifying that it is conjugate-linear.
• Turning kets into bras without conjugating coefficients.
• Treating a global phase and a relative phase as the same kind of phase.
• Assuming every mathematically possible scalar field gives the same physical quantum theory without extra interpretation.

Cross-Links

• Complex Numbers
• Vector Spaces and Dual Spaces
• Dirac Notation as Linear Algebra
• Linear Maps
• Inner Products
• Finite-Dimensional Hilbert Spaces
• Antiunitary Symmetries, First Look
• State Vectors
• Superposition and Relative Phase
• Antiunitary Symmetries

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Show that complex conjugation $K(z)=z^*$ is not complex-linear.

Solution

Complex linearity would require $K(iz)=iK(z)$. But

$$K(iz)=(iz)^*=-iz^*,$$

while

$$iK(z)=iz^*.$$

These are unequal for nonzero $z$, so $K$ is not complex-linear.

2. What are the complex and real dimensions of $\mathbb C^3$?

Solution

As a complex vector space, $\mathbb C^3$ has dimension $3$. As a real vector space, it has dimension $6$, because each complex coordinate contributes a real and imaginary part.

3. With the physics convention, what bra corresponds to

$$\lvert\psi\rangle = \alpha\lvert0\rangle+\beta\lvert1\rangle?$$

Solution

The corresponding bra is

$$\langle\psi\vert = \alpha^*\langle0\vert+\beta^*\langle1\vert.$$

The coefficients are conjugated because the vector-to-dual map is conjugate-linear.