Math Behind Quantum Computing

In short: Quantum computing is linear algebra over complex numbers: a state is a unit vector, multi-qubit systems are tensor products, gates are unitary matrices, and measurement turns squared amplitudes into probabilities via the Born rule.

Screening rounds for quantum roles rarely ask you to recite physics history. They ask you to multiply a small matrix by a small vector, decide whether a two-qubit state is entangled, or explain why a gate must be unitary. If you can do those four things on a whiteboard, you pass most first-round filters. This article gives you the minimum rigorous toolkit, with numbers you can check by hand. For a gentler warm-up on matrices, see our linear algebra for quantum computing primer.

1. States live in complex vector spaces

A classical bit is 0 or 1. A qubit is a vector in a two-dimensional complex vector space, written in Dirac notation as |ψ⟩ = α|0⟩ + β|1⟩, where |0⟩ = (1, 0)ᵀ and |1⟩ = (0, 1)ᵀ form the computational basis. The numbers α and β are complex amplitudes, and the normalisation condition is |α|² + |β|² = 1. That single constraint makes every valid qubit state a point on a sphere (the Bloch sphere) rather than anywhere in the plane.

Why complex numbers and not real ones? Because phase matters. The states |+⟩ = (|0⟩ + |1⟩)/√2 and |−⟩ = (|0⟩ − |1⟩)/√2 give identical measurement odds in the computational basis, yet they behave differently when interfered through a Hadamard gate. Complex amplitudes let a single number carry both a magnitude and a phase angle.

Key fact: a global phase (multiplying the whole state by e^{iθ}) is unobservable, but a relative phase between |0⟩ and |1⟩ is physically real and drives all interference.

Two more definitions interviewers expect. The inner product ⟨φ|ψ⟩ conjugates the first vector's entries before multiplying; this is why the bra ⟨φ| is a conjugate transpose, not a plain transpose. Two states are orthogonal when ⟨φ|ψ⟩ = 0, and orthogonal states are exactly the ones a measurement can distinguish perfectly.

2. Tensor products: why n qubits need 2ⁿ numbers

To describe two qubits you do not stack their vectors, you take the Kronecker (tensor) product. For |a⟩ = (a₀, a₁)ᵀ and |b⟩ = (b₀, b₁)ᵀ, |a⟩⊗|b⟩ = (a₀b₀, a₀b₁, a₁b₀, a₁b₁)ᵀ. A worked example: |+⟩⊗|0⟩ = (1/√2)(1, 0, 1, 0)ᵀ, which is (|00⟩ + |10⟩)/√2.

The growth is the whole story behind "quantum advantage" and "classical simulation is hard". Storing n qubits needs 2ⁿ complex amplitudes. At 16 bytes per complex128 amplitude:

This is why a laptop simulates about 30 qubits comfortably and why state-vector simulators on supercomputers top out in the mid-40s to 50s.

The second interview favourite: entanglement. A two-qubit state (c₀₀, c₀₁, c₁₀, c₁₁) is a product state only if c₀₀c₁₁ = c₀₁c₁₀. The Bell state (|00⟩ + |11⟩)/√2 has c₀₀c₁₁ = 1/2 but c₀₁c₁₀ = 0, so no pair of single-qubit vectors can produce it. Entangled states are the vectors the tensor product cannot reach by multiplication, yet they remain valid vectors in the 4-dimensional space.

Key fact: most vectors in the 2ⁿ-dimensional space are entangled. Product states are a tiny, low-dimensional corner of it.

3. Gates are unitary matrices

Quantum evolution is a linear map that must keep vectors normalised, otherwise probabilities stop summing to 1. A matrix U that preserves every vector's length satisfies U†U = I, where U† is the conjugate transpose. That is the definition of unitary. Consequences you should be able to state: columns are orthonormal, |det U| = 1, eigenvalues have magnitude 1, and U is always invertible, with inverse U†. Reversibility is why quantum gates cannot copy or erase information the way a classical AND gate does.

The Hadamard gate is H = (1/√2)[[1, 1],[1, −1]]. Check H†H: with H real and symmetric this is H², and (1/2)[[2, 0],[0, 2]] = I. Acting on |0⟩ it gives |+⟩; acting on |+⟩ it returns |0⟩ because the |1⟩ components cancel by destructive interference. The Pauli-X gate [[0,1],[1,0]] is the quantum NOT, and the phase gate S = diag(1, i) rotates the relative phase by 90 degrees.

For multi-qubit gates, tensor products apply again. H⊗H on two qubits is a 4x4 unitary built from the single-qubit blocks, while CNOT is a 4x4 permutation matrix that flips the second qubit when the first is 1. Applying H to the first qubit of |00⟩ and then CNOT produces exactly the Bell state above, the textbook circuit for creating entanglement.

4. The Born rule: amplitudes to probabilities

Measurement in the computational basis returns outcome i with probability P(i) = |⟨i|ψ⟩|². Take |ψ⟩ = (√3/2)|0⟩ + (i/2)|1⟩. Then P(0) = 3/4 and P(1) = |i/2|² = 1/4. Notice the factor i changed nothing in the probability, yet it would matter if another gate followed. After you observe an outcome, the state collapses to that basis vector, so repeating the measurement gives the same answer.

For Bell pairs, P(00) = |1/√2|² = 1/2, P(11) = 1/2, and the cross terms are zero. Measuring one qubit instantly fixes the other, but no usable signal travels, a point interviewers love to probe. You can also measure in another basis by first applying a unitary, for instance H before a Z measurement to read out the X basis.

The expected value of an observable M (a Hermitian matrix) is ⟨ψ|M|ψ⟩. For Pauli-Z with eigenvalues +1 and −1 this equals P(0) − P(1), which is 1/2 for our example. This formula is the engine of variational algorithms such as VQE, where a classical optimiser minimises that expectation value.

5. Putting it together: a screening-round walkthrough

Here is a typical question: "Apply H to both qubits of |00⟩ and give the probability of measuring 11." Work it as a tensor product. H|0⟩ = |+⟩ on each qubit, so the state is |+⟩⊗|+⟩ = (1/2)(1, 1, 1, 1)ᵀ. Each amplitude is 1/2, so every outcome has probability 1/4. The answer is 25 percent, and you showed it is a product state, hence unentangled.

A harder variant: "Is a given 2x2 matrix a valid gate?" Compute U†U. For [[1, 1],[1, 1]]/√2 you get (1/2)[[2, 2],[2, 2]], which is not I, so it is not unitary. It would collapse |0⟩ and |1⟩ onto the same vector, destroying reversibility.

• Check normalisation first: squared magnitudes must sum to 1.
• Build multi-qubit states with Kronecker products, ordering qubits consistently (Qiskit uses little-endian, textbooks big-endian).
• Apply Born's rule only at the end, and use the conjugate when squaring complex amplitudes.

Key fact: tools such as Qiskit's Statevector, Cirq and NumPy let you verify any of these by hand-calculation in a few lines, a good habit before an interview.

6. Where the maths meets silicon

The same unitary structure shapes real hardware. A superconducting transmon gate is a calibrated microwave pulse implementing a rotation, and errors show up as non-unitary leakage. Control electronics, cryo-CMOS and the classical FPGA stacks that schedule pulses are designed with mainstream EDA flows such as Cadence Virtuoso and Synopsys tools, so VLSI engineers increasingly meet this vocabulary. If you want to build on the chip side, ChipXpert's VLSI training programs cover the digital and analog fundamentals that quantum control hardware depends on. For the next step in the maths, read our guide to quantum gates and circuits explained.

Ready to practise these calculations with guided problems and mock screening rounds? Explore the quantum track in our CourseTron courses.

Frequently asked questions

Why must quantum gates be unitary?

Unitary matrices preserve vector length, so total probability stays 1, and they are invertible, so evolution is reversible. A non-unitary map would either leak probability or merge distinct states.

What is the difference between global and relative phase?

A global phase multiplies the whole state and cannot be detected. A relative phase between basis components changes how states interfere under later gates, so it is physically meaningful.

How do I tell if a two-qubit state is entangled?

Write the amplitudes as a 2x2 matrix [[c00, c01],[c10, c11]]. The state is a product state exactly when its determinant c00c11 minus c01c10 is zero; otherwise it is entangled.

Why does simulating quantum computers get hard so quickly?

The state needs 2ⁿ complex amplitudes. At 50 qubits that is roughly 16 petabytes in double precision, beyond ordinary machines.