I just want to know why actually Bell states are examples of maximally entangled states and significance of that "maximal" term. Is there anything for proving that?


2 Answers 2


It depends on what you want to take as your definition of maximally entangled. But, here's a good one:

Given a Bell state, I can convert it into any other two-qubit entangled state using only local operations and classical communication

Given that local operations and classical communication cannot increase entanglement, the possibility of performing $|\psi\rangle\rightarrow|\phi\rangle$ naturally induces a partial ordering of states being more or less entangled than each other, and the Bell states sit at the top of that tree.

Proof of my claim: let's say I start with $(|00\rangle+|11\rangle)/\sqrt{2}$. I apply a measurement, which comprises two operators $$ \left(\begin{array}{cc} \alpha & 0 \\ 0 & \beta \end{array}\right)\otimes I,\qquad \left(\begin{array}{cc} \beta & 0 \\ 0 & \alpha \end{array}\right)\otimes I. $$ If I get the second outcome, I apply $X$ on both qubits (this requires classical communication). Thus, whichever result I get, I have the output $$ \alpha|00\rangle+\beta|11\rangle. $$ If I choose $\alpha,\beta$ to be the two Schmidt coefficients of my target state, I'm done, up to local unitary rotations.

  • $\begingroup$ But by saying "cannot increase entanglement", you might suppose the amount of entanglement first. $\endgroup$
    – narip
    Sep 1, 2021 at 13:49
  • 2
    $\begingroup$ @narip I don't agree. I'm starting from a qualitative definition, that entanglement is something generated from the interaction between two remote parties, and cannot be generated locally. From there, my argument is telling you that there is a partial ordering, and makes it quantifiable. $\endgroup$
    – DaftWullie
    Sep 1, 2021 at 14:25
  • $\begingroup$ That's very nice! I didn't know that LOCC can reduce the amount of entanglement. But this is solely due to measurements, right? $\endgroup$ Sep 1, 2021 at 15:33
  • $\begingroup$ @WeatherReport Yes, exactly. Local unitaries preserve the amount of entanglement. Measurements, on average, reduce entanglement (although individual results can increase the entanglement) $\endgroup$
    – DaftWullie
    Sep 2, 2021 at 6:45
  • 1
    $\begingroup$ @NikitaNemkov You are perhaps thinking of projective measurements. But generalised measurements give a greater range of freedom. Consider $(|00\rangle+2|11\rangle)/\sqrt{5}$ and use two measurement operators $\text{diag}(1,1/2)$ and $\text{diag}(0,\sqrt{3}/2)$. The first gives a maximally entangled output, the second gives a fully separable output. You might like to look up "Entanglement of Pure States for a Single Copy" by Vidal. $\endgroup$
    – DaftWullie
    Sep 6, 2021 at 6:49

Entanglement is analogous to correlation.

Correlation over a property simply exists where simultaneous determination is not equivalent to individual determination of a property. What does it mean?

Say you have 2 coins - one black and one white. Let's say if you have a black coin you get a $+1$ colour value whereas with white you get a $-1$ colour value. If you randomly distribute these coins to two people $A$ and $B$ and one person looks at a coin, they will immediately know the colour of the other coin. How? Well because if you have a colour, say white, you can exactly determine what colour the other person has ( although you may not convey that to them ).

Continuing with our example, let's say $v_a$ is the value $A$ expects to gain and $v_b$ is the value $B$ expects to gain. Since the distribution was random, the average value of $v_a$ is $0$ and similarly, the average value of $v_b$ is $0$. What is the average of $v_a v_b$? It is just $-1$ because the product is always going to be $-1$. What does it mean?

This means the average value ($+1\ or\ -1$) of the property i.e. colour value is correlated. Why? Because the average of the product i.e. $-1$ ( simultaneous determination ) is not equivalent to the product of the averages i.e. $0$.

A qubit $q_i$ is said to be entangled with qubit $q_j$ if determining the state of $q_i$ lets us know some information about the state of $q_j$. Maximal entanglement here means that determining the state of one qubit lets you completely determine the state of the other qubit. Moreover, if you only try to measure a single state, you would randomly get either a value of $0$ or $1$ and no useful information about the individual systems.

The four bell states are - $$ | \psi_1 \rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}\\ | \psi_2 \rangle = \frac{|01\rangle + |10\rangle}{\sqrt{2}}\\ | \psi_3 \rangle = \frac{|01\rangle - |10\rangle}{\sqrt{2}}\\ | \psi_4 \rangle = \frac{|00\rangle - |11\rangle}{\sqrt{2}}$$

Since the correlation between states of the two qubits is maximum, we say that the states are maximally entangled and simple visual proof for this is that for any Bell State, measuring one qubit completely determines what the state of the other qubit will be!

If you carefully see, in any of the four states, if you measure let's say qubit 1, you have a complete knowledge of what the state of qubit 2 is going to be after your measurement. Eg. in $|\psi_2 \rangle$ you see a measurement of qubit 1 completely determines the state of qubit 2 i.e. if qubit 1 is measured as $0$, qubit 2 would surely be measured as $1$ and vice versa.

Hope it helps!

  • $\begingroup$ Thank you for your valuable answer $\endgroup$
    – user18115
    Sep 2, 2021 at 16:52
  • $\begingroup$ Good answer and I like it! $\endgroup$ Apr 7, 2022 at 7:52
  • $\begingroup$ @gIS Moderators, I'm sorry I think someone has hacked my account and answered some questions using ChatGPT. I'm really sorry for this unhealthy discussions. I'm attemoting to delete this account. – $\endgroup$
    – user18115
    Mar 12, 2023 at 5:31

Your Answer

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