DeepPractise
DeepPractise

Hadamard Gate

Track: Quantum Gates & Circuits · Difficulty: Beginner · Est: 14 min

Hadamard Gate

Overview

The Hadamard gate (H) is one of the most important single-qubit gates. It is the cleanest way to:

  • create superposition from a basis state,
  • undo that superposition, and
  • switch between measurement bases (Z-basis ↔ X-basis).

Even when we avoid circuits, H is central because it teaches the core idea behind quantum behavior:

  • probabilities come from amplitudes, and amplitudes can interfere.

Intuition

On the Bloch sphere:

  • the Z-basis states 0|0\rangle and 1|1\rangle are the north and south poles (+zz and −zz).
  • the X-basis states are +=12(0+1),=12(01).|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle),\qquad | - \rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle). These lie on the equator at +xx and −xx.

The Hadamard gate maps poles to equator points:

  • 0+|0\rangle \mapsto |+\rangle
  • 1|1\rangle \mapsto | - \rangle

So H is a geometric “tilt” that changes which basis looks like “up/down.”

One more crucial intuition:

  • Applying H twice gets you back to where you started.

So H is its own inverse. That “do it again to undo it” property makes H a natural tool for creating and then later removing superposition.

Formal Description

Action on basis states

The defining behavior is

H0=+=12(0+1),H|0\rangle = |+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle), H1==12(01).H|1\rangle = | - \rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle).

This already tells you what H does to any superposition by linearity.

Matrix form (with meaning)

In the computational basis, H is represented by

H=12(1111).H = \tfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}.

Let’s interpret the entries rather than “dumping a matrix”:

  • The factor 12\tfrac{1}{\sqrt{2}} keeps the output normalized.
  • The first column describes what happens to 0|0\rangle.
  • The second column describes what happens to 1|1\rangle.

You can check this by multiplying:

  • 0(10)|0\rangle \equiv \begin{pmatrix}1\\0\end{pmatrix} gives the first column: H0=12(11)=12(0+1)=+.H|0\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |+\rangle.
  • 1(01)|1\rangle \equiv \begin{pmatrix}0\\1\end{pmatrix} gives the second column: H1=12(11)=12(01)=.H|1\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = | - \rangle.

A key identity (stated conceptually) is:

H2=I.H^2 = I.

Meaning: applying H twice returns the original state.

Worked Example

Start with 0|0\rangle.

  1. Apply H:
0  H  +=12(0+1).|0\rangle \xrightarrow{\;H\;} |+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle).

Geometrically: north pole (+zz) moves to +xx on the equator.

  1. Apply H again:
+  H  0.|+\rangle \xrightarrow{\;H\;} |0\rangle.

So H both creates and undoes superposition.

Interpretation:

  • H is a basis change between the Z and X viewpoints.
  • “Superposition” is not mystical; it is often just “a basis state in a different basis.”

Turtle Tip

Turtle Tip

If you ever feel lost, translate H into one sentence: H converts ‘definitely 0/1 in Z’ into ‘definitely +/− in X’, and vice versa.

Common Pitfalls

Common Pitfalls
  • Don’t say “H makes a random bit.” H makes a deterministic state with two amplitudes; randomness appears only after measurement.
  • Don’t forget that +|+\rangle and | - \rangle are as “definite” as 0|0\rangle and 1|1\rangle—just in a different basis.

Quick Check

Quick Check
  1. What are H0H|0\rangle and H1H|1\rangle?
  2. What does H2=IH^2 = I mean in plain language?

What’s Next

H makes superposition and changes basis. Next we study gates that mainly change phase rather than probabilities: the phase gates S and T, which rotate around the zz axis by smaller angles than Z.