The Bell state $|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$ is an entangled state. But why is that the case? How do I mathematically prove that?


4 Answers 4


A two qudit pure state is separable if and only if it can be written in the form $$|\Psi\rangle=|\psi\rangle|\phi\rangle$$ for arbitrary single qudit states $|\psi\rangle$ and $|\phi\rangle$. Otherwise, it is entangled.

To determine if the pure state is entangled, one could try a brute force method of attempting to find satisfying states $|\psi\rangle$ and $|\phi\rangle$, as in this answer. This is inelegant, and hard work in the general case. A more straightforward way to prove whether this pure state is entangled is the calculate the reduced density matrix $\rho$ for one of the qudits, i.e. by tracing out the other. The state is separable if and only if $\rho$ has rank 1. Otherwise it is entangled. Mathematically, you can test the rank condition simply by evaluating $\text{Tr}(\rho^2)$. The original state is separable if and only if this value is 1. Otherwise the state is entangled.

For example, imagine one has a pure separable state $|\Psi\rangle=|\psi\rangle|\phi\rangle$. The reduced density matrix on $A$ is $$ \rho_A=\text{Tr}_B(|\Psi\rangle\langle\Psi|)=|\psi\rangle\langle\psi|, $$ and $$ \text{Tr}(\rho_A^2)=\text{Tr}(|\psi\rangle\langle\psi|\cdot |\psi\rangle\langle\psi|)=\text{Tr}(|\psi\rangle\langle\psi|)=1. $$ Thus, we have a separable state.

Meanwhile, if we take $|\Psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$, then $$ \rho_A=\text{Tr}_B(|\Psi\rangle\langle\Psi|)=\frac12\left(|0\rangle\langle 0|+|1\rangle\langle 1|\right)=\frac12\mathbb{I} $$ and $$ \text{Tr}(\rho_A^2)=\frac14\text{Tr}(\mathbb{I}\cdot\mathbb{I})=\frac12 $$ Since this value is not 1, we have an entangled state.

If you wish to know about detecting entanglement in mixed states (not pure states), this is less straightforward, but for two qubits there is a necessary and sufficient condition for separability: positivity under the partial transpose operation.

  • 1
    $\begingroup$ +1 This is a much more elegant method compared to the brute force algorithm. $\endgroup$ Commented Jun 8, 2018 at 10:03
  • $\begingroup$ What are $A$ and $B$? Are these just the qudits themselves? $\endgroup$
    – Urðr
    Commented Jul 10, 2019 at 9:08
  • 1
    $\begingroup$ @Dohleman Yes, they're just labels for the two parts of the system, one part held by A (Alice), and the other by B (Bob). In this case it's the two qudits. $\endgroup$
    – DaftWullie
    Commented Jul 15, 2019 at 7:21

Actually an even simpler way is as follows (reusing @nbro's notations). We have:

\begin{align} |\Phi^{+}\rangle &= |a\rangle \otimes |b\rangle \\ &= \left( \alpha |0\rangle + \beta |1\rangle \right) \otimes \left( \gamma |0\rangle + \lambda |1\rangle \right) \end{align}

We can now simply apply the distributive property to obtain

\begin{align} |\Phi^{+}\rangle &= \cdots \\ &= \left( \alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle \right) \end{align}

Now if we multiply the coefficients of $|00\rangle $ and $|11\rangle $, we get $\alpha \beta \gamma \lambda$.

Also, if we multiply the coefficients of $|01\rangle $ and $ |10\rangle $, we get $\alpha \beta \gamma \lambda$

If these two are equal, then the qubits are in a product state. Else, it automatically means the qubits are in an entangled state.

For the Bell state above, the first product = $1/2$ and the second product = 0. Since they are unequal they are in an entangled state.

I find this is the quickest and easiest way to tell if a pair of qubits are entangled or not.

  • $\begingroup$ That works for a pure two qubit state. You might want to add a proof for your claim that if the two products are equal => the state is a product state? $\endgroup$
    – M. Stern
    Commented Oct 10, 2021 at 17:11
  • $\begingroup$ hmm... I am not sure what exactly you are looking for. Basically, starting out with arbitrary variables α,β,γ, and λ, doesn't this already show they need to be equal to be in a product state... $\endgroup$
    – codester
    Commented Oct 11, 2021 at 3:02
  • 2
    $\begingroup$ I mean the other direction: if the product of the coefficients in front of |00> and |11> equals the product of coefficients in front of |01> and |10>, then the state is a product state. $\endgroup$
    – M. Stern
    Commented Oct 11, 2021 at 17:59
  • $\begingroup$ A nice explanation can be found in this answer. To conclude, if we are given an equation of form |Φ>=a|00>+b|01>+c|10>+d|11> then |Φ> is a product state iff ad=bc. $\endgroup$
    – manud99
    Commented Oct 15, 2023 at 6:57

Take a look at "entanglement of formation". It's a nice metric for how much is a state entangled. the "units" are (a log of) how many bell states required to get into this state. See here for more info, and also here.


from qiskit.quantum_info import entanglement_of_formation
import numpy as np
li = np.sqrt(0.5) #shortcut for 1/sqrt(2)

Bell state:

bell = [li, 0, 0, li]
#returns 0.9999999999999999 (practically 1...)

|00> state:

just00 = [1, 0, 0, 0] 
#returns 0.0

Not really Bell (|+0>):

plus0 = [li, li, 0, 0]
#returns 1.6017132519074586e-16 (practically zero...)
  • $\begingroup$ When you think about the chosen metric, it amusingly follows... measuring Bell states in Bells (or dB :)) $\endgroup$
    – learn2
    Commented Dec 28, 2021 at 7:03

There are several ways. Easiest of the is:

  1. Find $\rho = |\Psi\rangle\langle\Psi|$
  2. Find $\rho^2$
  3. Check if $\rho^2 = \rho$. If yes, it is a pure state and entangled. This state $|\Psi\rangle$ can not be written as $|\psi\rangle\otimes|\phi\rangle$ form.

Another way is to find Tr($\rho^2$). If it $1$, then the state is entangled.

Third way would be checking the Schmidt decomposition method.

  • 1
    $\begingroup$ Pure state can be either entangled or separable. Pure and entangled are not synonyms. $\endgroup$ Commented Nov 8, 2022 at 6:55
  • 1
    $\begingroup$ You are right. I have used "i.e." instead of "and/also". Thanks for correcting me. I'm going to modify it :) $\endgroup$ Commented Nov 10, 2022 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.