Mathematical representation of two-entangled q-bits?

[Emrissa]

Dear all,I have four questions. Hopefully, someone can answer. Thank you :)

1.
A qubit is described as a two-orthogonal basis state. How about two entangled qubits?

2.
What is the actual reason for a qubit cannot be cloned/copied?
Is it because without knowing the value of the complex number there is no way to read the actual state of the qubit? Besides, the only way is to measure the qubit. However this process collapses the coherence superposition state of a qubit.

3.
A single qubit is described as a Two-Dimensional Hilbert Space.
Are two entangled qubits also referred to as a Four-Dimensional Hilbert Space? The only difference is that their quantum states cannot be described as independent entities.

4.
May I know which one is the correct mathematical representation of two entangled qubits:
*Please refer to the attached file.

 

[PeterDonis]

A qubit is described as a two-orthogonal basis state.

Not quite. A qubit is a very simple quantum system whose Hilbert space is two-dimensional. That means an orthonormal basis for the Hilbert space will consist of two vectors. But there are an infinite number of possible choices of basis.

How about two entangled qubits?

Whether entangled or not, two qubits are a quantum system whose Hilbert space is four-dimensional (it is the tensor product of two two-dimensional one-qubit Hilbert spaces, modulo some technicalities). The Hilbert space is the same regardless of whether the qubits are entangled or not.

What determines whether the two qubits are entangled is whether the particular state vector of the two-qubit system in the two-qubit Hilbert space can be written as a product of two one-qubit states (in which case the qubits are not entangled), or not (in which case the qubits are entangled).

What is the actual reason for a qubit cannot be cloned/copied?

Not just a qubit, but any quantum system, cannot be cloned.

Is it because without knowing the value of the complex number there is no way to read the actual state of the qubit?

Sort of. See below.

Besides, the only way is to measure the qubit.

That's not the only possible way you could clone a qubit. The other possible way would be to realize a unitary operation (what quantum computing people would call a "quantum gate") that manipulated a second qubit to be in the same state as the first qubit (the one to be copied), without making a measurement. But neither of these ways work; that is the content of the no-cloning theorem.

However this process collapses the coherence superposition state of a qubit.

Not necessarily. If the qubit is already in an eigenstate of the measurement operator, measuring it won't change its state. However, if you know the qubit is in an eigenstate of a known measurement operator, you don't have to measure it to prepare another qubit in the same state. The no-cloning theorem doesn't apply to this kind of thing; it only applies to attempts to clone a qubit when you don't already know what state it's in. So a better way to restate what you are trying to say here is that, if you don't know what state a qubit is in, you don't know whether you changed its state or not if you measure it, so you can't use a measurement to clone its state.

A single qubit is described as a Two-Dimensional Hilbert Space.
Are two entangled qubits also referred to as a Four-Dimensional Hilbert Space?

Yes to both. See above.

Please refer to the attached file

Posting equations in attachments is not allowed. Please use the PF LaTeX feature to post equations directly. You will find a "LaTeX Guide" link to the lower left of each post window.

 

[jedishrfu]

Wikipedia has a comprehensive writeup on the qubit.

They mention that an n-qubit system is basicallya $2^n$ Hilbert space.

https://en.wikipedia.org/wiki/Qubit

 

[Emrissa]

Not quite. A qubit is a very simple quantum system whose Hilbert space is two-dimensional. That means an orthonormal basis for the Hilbert space will consist of two vectors. But there are an infinite number of possible choices of basis.Whether entangled or not, two qubits are a quantum system whose Hilbert space is four-dimensional (it is the tensor product of two two-dimensional one-qubit Hilbert spaces, modulo some technicalities). The Hilbert space is the same regardless of whether the qubits are entangled or not.

What determines whether the two qubits are entangled is whether the particular state vector of the two-qubit system in the two-qubit Hilbert space can be written as a product of two one-qubit states (in which case the qubits are not entangled), or not (in which case the qubits are entangled).Not just a qubit, but any quantum system, cannot be cloned.Sort of. See below.That's not the only possible way you could clone a qubit. The other possible way would be to realize a unitary operation (what quantum computing people would call a "quantum gate") that manipulated a second qubit to be in the same state as the first qubit (the one to be copied), without making a measurement. But neither of these ways work; that is the content of the no-cloning theorem.Not necessarily. If the qubit is already in an eigenstate of the measurement operator, measuring it won't change its state. However, if you know the qubit is in an eigenstate of a known measurement operator, you don't have to measure it to prepare another qubit in the same state. The no-cloning theorem doesn't apply to this kind of thing; it only applies to attempts to clone a qubit when you don't already know what state it's in. So a better way to restate what you are trying to say here is that, if you don't know what state a qubit is in, you don't know whether you changed its state or not if you measure it, so you can't use a measurement to clone its state.Yes to both. See above.Posting equations in attachments is not allowed. Please use the PF LaTeX feature to post equations directly. You will find a "LaTeX Guide" link to the lower left of each post window.

 

[PeterDonis]

@Emrissa, there is no content in your post #4, just a long quote of my previous post.

 

[Emrissa]

@Emrissa, there is no content in your post #4, just a long quote of my previous post.

Dear @PeterDonis,
Thank you very much for the reply, that's helpful.
I'm a first-time user... Now I'm investigating how to write the equation directly in this post. How about this?

 

[PeterDonis]

How about this?

That's an image, which is still not acceptable. As I said, you need to use the PF LaTeX feature to post equations, and there is a "LaTeX Guide" link at the bottom left of each post window that will help you with that. Text that isn't equations can just be written directly in the post, as you have been doing.

 

[PeterDonis]

@Emrissa, as an example of how to use the PF LaTeX feature, I'll post your two equations, first showing the LaTeX code and the other text just as you would type it in the post window, and then the result when that code is posted.

Which one is the correct mathematical representation of two entangled qubits?

1.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{01} + \gamma \ket{10} + \delta \ket{11}
$$

or

2.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{11}
$$

Which one is the correct mathematical representation of two entangled qubits?

1.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{01} + \gamma \ket{10} + \delta \ket{11}
$$

2.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{11}
$$

 

[Emrissa]

@Emrissa, as an example of how to use the PF LaTeX feature, I'll post your two equations, first showing the LaTeX code and the other text just as you would type it in the post window, and then the result when that code is posted.

Which one is the correct mathematical representation of two entangled qubits?

1.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{01} + \gamma \ket{10} + \delta \ket{11}
$$

or

2.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{11}
$$

Which one is the correct mathematical representation of two entangled qubits?

1.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{01} + \gamma \ket{10} + \delta \ket{11}
$$

2.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{11}
$$

Dear Sir,
Thank you for your guidance

 

[PeterDonis]

which one is the correct mathematical representation of two entangled qubits

The answer to this is that state #1 is not entangled and state #2 is. However, state #2 is not the only possible entangled state of two qubits. So there is no such thing as "the" correct mathematical representation of two entangled qubits.

To see that state #1 is not entangled, consider:

$$
\ket{\Psi_{AB}} = \left( a \ket{0} + b \ket{1} \right) \left( c \ket{0} + d \ket{1} \right)
$$

Multiplying out the factors gives:

$$
\ket{\Psi_{AB}} = ac \ket{00} + ad \ket{01} + bc \ket{10} + bd \ket{11}
$$

Since $a$, $b$, $c$, and $d$ were all arbitrary coefficients, we simply set $\alpha = ac$, $\beta = ad$, $\gamma = bc$, and $\delta = bd$, and we have a factorization of the original state #1. Since the state can be factored into a product of one-qubit states, it is not entangled.

 

[Emrissa]

The answer to this is that state #1 is not entangled and state #2 is. However, state #2 is not the only possible entangled state of two qubits. So there is no such thing as "the" correct mathematical representation of two entangled qubits.

To see that state #1 is not entangled, consider:

$$
\ket{\Psi_{AB}} = \left( a \ket{0} + b \ket{1} \right) \left( c \ket{0} + d \ket{1} \right)
$$

Multiplying out the factors gives:

$$
\ket{\Psi_{AB}} = ac \ket{00} + ad \ket{01} + bc \ket{10} + bd \ket{11}
$$

Since $a$, $b$, $c$, and $d$ were all arbitrary coefficients, we simply set $\alpha = ac$, $\beta = ad$, $\gamma = bc$, and $\delta = bd$, and we have a factorization of the original state #1. Since the state can be factored into a product of one-qubit states, it is not entangled.

"However, state #2 is not the only possible entangled state of two qubits"

This means, besides state #2, the other possible entangled state of two qubits is state #3?

2.
$$
\ket{\Psi_{AB}} = \alpha \ket{00} + \beta \ket{11}
$$

3.
$$
\ket{\Psi_{AB}} = \alpha \ket{01} + \beta \ket{10}
$$

 

[PeterDonis]

This means, besides state #2, the other possible entangled state of two qubits is state #3?

That is another one, but there are still others. In fact, infinitely many. There are four commonly used two-qubit entangled states that form a basis of the two-qubit Hilbert space, the Bell states:

https://en.wikipedia.org/wiki/Bell_state

However, there are also an infinite number of states that are linear combinations of multiple Bell states and are still entangled.

 

[Emrissa]

That is another one, but there are still others. In fact, infinitely many. There are four commonly used two-qubit entangled states that form a basis of the two-qubit Hilbert space, the Bell states:

https://en.wikipedia.org/wiki/Bell_state

However, there are also an infinite number of states that are linear combinations of multiple Bell states and are still entangled.

@PeterDonis, thank you so much

 

[Emrissa]

That is another one, but there are still others. In fact, infinitely many. There are four commonly used two-qubit entangled states that form a basis of the two-qubit Hilbert space, the Bell states:

https://en.wikipedia.org/wiki/Bell_state

However, there are also an infinite number of states that are linear combinations of multiple Bell states and are still entangled.

Regarding to the superposition state --

Besides coexist in state

$$
\ket{0} AND \ket{1}
$$

Qubit can be in any state pointing the surface of the Bloch Sphere? Is that true?

 

[vanhees71]

Dear @PeterDonis,
Thank you very much for the reply, that's helpful.
I'm a first-time user... Now I'm investigating how to write the equation directly in this post. How about this?

View attachment 317880

The most general two-qubit state is given by 1. A two-qubit state is called "entangled" if it cannot be written as a product of two single-qubit states (all this for distinguishable qubits; for photons and other indistinguishable particles it's easier to use the annihilation and creation operators acting on the vacuum state to define the many-body states to automatically implement the Bose-Einstein and Fermi-Dirac commutation/anti-commutation relations).

You state No. 2. is an example for an entangled state, if neither $\alpha$ nor $\beta$ vanishes.

Concerning LaTeX, see the LaTeX Guide link below the web form used to type postings:
https://www.physicsforums.com/help/latexhelp/

 

[vanhees71]

Regarding to the superposition state --

Besides coexist in state

$$
\ket{0} AND \ket{1}
$$

Qubit can be in any state pointing the surface of the Bloch Sphere? Is that true?

That doesn't make sense. You have to define, what you mean by "AND".

 

[Emrissa]

That doesn't make sense. You have to define, what you mean by "AND".

AND means both exist simultaneously.

 

[vanhees71]

That doesn't make sense at all. $|0 \rangle$ and $|1 \rangle$ are orthogonal to each other, i.e., if the qubit is prepared in the state $|0 \rangle$, then the probability to be found in $|1 \rangle$ is $|\langle 1|0 \rangle|^2=0$, i.e., it is with certainty never found to be in $|1 \rangle$.

For get about all the popular-science lingo about "a qubit being at the same time in two states". What they usually refer to is that the qubit can be in any superposition of these two states,
$$|\psi \rangle=\alpha |0 \rangle + \beta |1 \rangle, \quad \langle \psi|\psi \rangle=|\alpha|^2+|\beta|^2=1.$$
The only meaning of this is that the probability to find the particle in either of the two states is
$$P(0)=|\langle 0|\psi \rangle|^2=|\alpha|^2, \quad P(1)=|\langle 1|\psi \rangle|^2=|\beta|^2.$$
This is at the heart of understanding quantum theory, and the two-dimensional Hilbert space (realized in nature by the spin of spin-1/2 particles or the polarization states of photons), is the most simple example. For a good approach to QT using this approach, see

J. J. Sakurai and S. Tuan, Modern Quantum Mechanics, Revised Edition, Addison Wesley (1993).

 

[Emrissa]

Does this mean superposition gives extra merit to qubit in terms of the data capacity it can encode? Because the classical bit can only represent one of two states (1 OR 0) at a time.

 

[PeterDonis]

Qubit can be in any state pointing the surface of the Bloch Sphere? Is that true?

Yes, the Bloch sphere is a valid representation of the single qubit Hilbert space.

 

[PeroK]

That is another one, but there are still others. In fact, infinitely many. There are four commonly used two-qubit entangled states that form a basis of the two-qubit Hilbert space, the Bell states:

https://en.wikipedia.org/wiki/Bell_state

However, there are also an infinite number of states that are linear combinations of multiple Bell states and are still entangled.

There is a homework problem here on the conditions for the coefficients for entanglement:

https://www.physicsforums.com/threa...unentangled-under-a-certain-condition.994430/

 

[vanhees71]

Yes, the Bloch sphere is a valid representation of the single qubit Hilbert space.

The Bloch sphere is more general. It's a representation of all possible pure and mixed states.

 

[PeterDonis]

The Bloch sphere is more general. It's a representation of all possible pure and mixed states.

The Bloch sphere itself is a 2-sphere that represents all possible pure states of a qubit. The 3-ball consisting of the 2-sphere and its interior represents all possible pure and mixed states of a qubit (the states on the surface of the 2-sphere are pure, the states in the interior of the 3-ball are mixed).