I was reading some material on QC online and I found some material that explains how to show that Bell states are orthonormal but without details.

I understand that we need to check $\langle state1|state2\rangle = 0$ but I'm not really sure how to calculate that with Dirac notation.

In the link above he somehow omitted the bra and the kets and just used the coefficients directly? I'm not sure I understand... Could someone please clarify a bit more?


3 Answers 3


First, note that $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and therefore $\langle 0 |1\rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix}= 0 $ and similarly, $ \langle 1|0\rangle = 0$. Also note that $|00\rangle = |0\rangle \otimes |0\rangle$, $|01\rangle = |0\rangle \otimes |1\rangle$ etc. Hence, $\langle 00| 11\rangle = 0$, $\langle 00| 10 \rangle = 0$, etc. But $\langle 00| 00\rangle = 1$ and $\langle 11 | 11\rangle = 1$ should be obvious.

The four Bell states are:

$|\Phi^+ \rangle = \dfrac{1}{\sqrt{2}}\big(|00\rangle + |11\rangle \big) $
$|\Phi^- \rangle = \dfrac{1}{\sqrt{2}}\big(|00\rangle - |11\rangle \big) $
$|\Psi^+ \rangle = \dfrac{1}{\sqrt{2}}\big(|01\rangle + |10 \rangle \big) $
$|\Psi^- \rangle = \dfrac{1}{\sqrt{2}}\big(|01\rangle - |10 \rangle \big) $

So therefore we have:

\begin{align} \langle \Phi^- |\Phi^+ \rangle &= \dfrac{1}{\sqrt{2}}\bigg(\langle 00 | - \langle 11|\bigg) \dfrac{1}{\sqrt{2}}\bigg(|00\rangle + |11\rangle \bigg) \\ &= \dfrac{1}{2} \bigg( \langle 00 |00\rangle + \langle 00 | 11\rangle - \langle 11 | 00\rangle - \langle 11| 11 \rangle \bigg) \\ &= \dfrac{1}{2}\bigg( 1 + 0 - 0 - 1 \bigg) \\ &= 0 \end{align}


\begin{align} \langle \Phi^- |\Psi^+ \rangle &= \dfrac{1}{\sqrt{2}}\bigg(\langle 00 | - \langle 11|\bigg) \dfrac{1}{\sqrt{2}}\bigg(|01\rangle + |10 \rangle \bigg)\\ &= \dfrac{1}{2} \bigg( \langle 00 |01\rangle + \langle 00 | 10\rangle - \langle 11 | 01\rangle - \langle 11| 10 \rangle \bigg) \\ &= \dfrac{1}{2}\bigg( 0 + 0 - 0 - 0 \bigg) \\ &= 0 \end{align}

And you can show that this is true for the others.

  • $\begingroup$ Thanks a lot for giving the details! This is exactly what I needed :) $\endgroup$
    – user206904
    Dec 30, 2020 at 10:50

The most basic but laborious way of checking that Bell states are orthonormal is to carry out the calculations for all sixteen inner products such as $\langle\Phi^+|\Psi^-\rangle$.

One way to do this is to switch from Dirac notation to standard linear algebra by replacing the kets and bras with appropriate column and row vectors. After this conversion you employ the formula for the complex dot product. Alternatively, you can perform the entire calculation in Dirac notation using known orthonormality relations such as those for the computational basis.

I suppose you were expecting the notes to use the letter method, but they employed the former. The methods are equivalent and should give you the same result.

The task can be made a little less laborious and notation more compact by exploiting the fact that the components of all four Bell states make up the columns of the matrix

$$ U = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{pmatrix}. $$

This way the task of verifying orthonormality of the Bell states becomes the task of checking that $U$ is unitary, i.e. that $U^\dagger U = I$.

In some contexts (e.g. Nielsen & Chuang, section 1.3.6, p. 26) Bell states are introduced as the output of the quantum circuit

$$ CNOT \circ (H \otimes I) \,|x\rangle|y\rangle\tag1 $$

on the four computational basis states $|0\rangle|0\rangle, |0\rangle|1\rangle, |1\rangle|0\rangle$ and $|1\rangle|1\rangle$ as inputs. This way of defining the states makes the task of checking their orthonormality the easiest: the conclusion follows from the facts that the computational basis states are orthonormal and that unitaries (such as our quantum circuit $(1)$) preserve inner products.

Finally, Bell states can be represented as simple tensor networks. In this view, $|\Phi^+\rangle$ represents a "cup" and the other three Bell states differ from it by a Pauli operator. Inner product of $A$ and $B$ links up the two open ports of $A$ with the two open ports of $B$ into a closed loop which represents the trace of the product of the operators on it. When the two Bell states in the inner product are the same then the Pauli operators cancel and the result is the trace of $I/2$, i.e. one. Otherwise, it equals the trace of a Pauli operator, i.e. zero.

  • $\begingroup$ Ah, now I see I provided same answer as you. +1 $\endgroup$ Dec 30, 2020 at 9:29

To avoid long computation of all pair-wise inner products in Bell basis, you can think as follow.

Switching from computational basis to the Bell basis is done by transformation $$ CNOT(H \otimes I) $$

As any operation on quantum computer (measurement and reset being exceptions), this operation is unitary. A unitary operation preserve angles among vectors and their length (norm).

Since computational basis is orthonormal (for two qubits, it is simply 4x4 unit matrix), a unitary operation on this basis has to produce again othonormal basis as all angles and vector lenghts are preserved.


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