The bracket of two states measures how much they overlap. A state has overlap one with itself when it is normalised, and overlap zero with a state it can be told apart from with certainty. In between, the overlap is an amplitude: its modulus squared is the probability that a system prepared in one state passes a test for the other. The Cauchy–Schwarz inequality is the guarantee that this never exceeds one.

For arrows, the dot product measures how much one points along another. The complex inner product is the same idea with conjugation built in, so that the length of an arrow always comes out real and positive.

The proof of the Cauchy–Schwarz inequality in a picture. The dot is the part of $\psi$ along $\phi$; what is left, $\chi$, is at right angles to $\phi$. Since $\chi$ has a length of at least zero, the part along $\phi$ can be no longer than $\psi$ itself.

The inner product of two vectors is their bracket $\braket{\phi|\psi}$. It is linear in the ket, conjugate-linear in the bra, conjugate-symmetric, and positive on every nonzero vector. The norm, or length, of a vector is $\lVert\psi\rVert=\sqrt{\braket{\psi|\psi}}$; a state is normalised when its norm is 1.

• Two vectors are orthogonal when $\braket{\phi|\psi}=0$. Orthogonal states are the ones a single measurement can tell apart with certainty.
• Normalising: $\ket{\psi}/\lVert\psi\rVert$ has norm 1 and describes the same state.
• For normalised states, $\mathrm{\mid }⟨\varphi \mathrm{\mid }$ is the probability that a system in is found to be in . It is 1 when the states are the same up to phase and 0 when they are orthogonal.

Split $\ket{\psi}$ into a part along $\ket{\phi}$ and a remainder orthogonal to it. Pythagoras makes the squared length of $\ket{\psi}$ the sum of the two parts' squared lengths, and dropping the remainder, which is at least zero, gives the inequality. Equality holds exactly when there is no remainder, that is when $\ket{\psi}$ is a multiple of .

For $\ket{\phi}\neq0$, subtract from $\ket{\psi}$ its component along $\ket{\phi}$. (If both sides are zero.)

The remainder is orthogonal to $\ket{\phi}$: that is what the choice of $c$ was for.

Since $\ket{\psi}=\ket{\chi}+c\ket{\phi}$ with the two parts orthogonal, the cross terms vanish.

Drop $\braket{\chi|\chi}\ge0$ and substitute $|c|^{2}\braket{\phi|\phi}=|\braket{\phi|\psi}|^{2}/\braket{\phi|\phi}$.

Multiply through by the positive number $\braket{\phi|\phi}$. Equality needs $\ket{\chi}=0$, that is $\ket{\psi}$ a multiple of $$.

The length of a vector is the square root of its inner product with itself, the sum of the moduli squared of its components.

The inner product conjugates the components of the left-hand vector. Leaving the conjugation out gives a wrong answer that can look very reasonable.

Divide a vector by its norm to get a normalised vector describing the same state.

For normalised states, the modulus squared of the inner product is the probability that a system prepared in one is found in the other.

Great work! You now know how to:

• compute norms and normalise vectors
• test orthogonality without forgetting the conjugate
• read the modulus squared of an overlap as a probability
• prove the Cauchy–Schwarz inequality and say when it is an equality