Why is entanglement necessary in a quantum computer?

[Trikenstein]

Intuitively, I find that more qbits and superposition of the quantum state could make sense as factor of the computation power of a quantum computer.

In Youtube videos or blogs about quantum computers. Entanglement is mentioned as an important requirement. But without a high level explanation of why and how.

Why are entangled qbits necessary?

What would happen to a quantum computer if it has a lots of qbits but without using any entangled qbits?

 

[PeroK]

What would happen to a quantum computer if it has a lots of qbits but without using any entangled qbits?

It would be a non-quantum computer!

 

[Hill]

"Entangled states are a uniquely quantum phenomenon; they have no classical counterpart. Most states in a multiple-qubit system are entangled states; they are what fills the vast quantum state spaces. The impossibility of efficiently simulating the behavior of entangled states on classical computers suggested to Feynman, Manin, and others that it might be possible to use these quantum behaviors to compute more efficiently, leading to the development of the field of quantum computation."

Rieffel, Eleanor G.; Polak, Wolfgang H.. Quantum Computing: A Gentle Introduction (Scientific and Engineering Computation) (p. 31). The MIT Press.

 

[gentzen]

In Youtube videos or blogs about quantum computers. Entanglement is mentioned as an important requirement. Why is it and how?

https://www.quantamagazine.org/the-quest-to-quantify-quantumness-20231019/

In 2002, Jozsa helped work out a simple process for using a classical computer to simulate a quantum “circuit,” which is a specific series of operations performed on qubits. If you gave the classical program some initial arrangement of qubits, it would predict their final arrangement, after they had been through the quantum circuit. Jozsa proved that, so long as his algorithm simulated a circuit that didn’t entangle qubits, it could handle larger and larger numbers of qubits without taking an exponentially longer time to run.

In other words, he showed that an entanglement-free quantum circuit was easy to simulate on a classical computer. In a computational sense, the circuit wasn’t intrinsically quantum. The collection of all such non-entangling circuits (or, equivalently, all arrangements of qubits that might come out of these non-entangling circuits) formed something of a classically simulable island in a vast quantum sea.

What would happen to a quantum computer if it has a lots of qbits but without using any entangled qbits?

It could be simulated by a classical computer without taking an exponentially longer time to run.

 

[Trikenstein]

Thanks you Mr Hill and Gentzen. Is it correct to extract from your answers that:

• Thanks to entanglement, the qbits can have a significantly larger number of states
• Without entanglement, the observable states are just a small subset ("island in a vast quantum sea" from in gentzen's answer)

 

[Hill]

Thanks you Mr Hill and Gentzen. Is it correct to extract from your answers that:

• Thanks to entanglement, the qbits can have a significantly larger number of states
• Without entanglement, the observable states are just a small subset ("island in a vast quantum sea" from in gentzen's answer)

Entanglement allows system of qubits to have a much larger set of states than combinations of individual qubit states would allow. But there is more to it. It allows for a new behavior where a state of entire system of entangled qubits changes at once, not by changing states of individual qubits.

 

[Trikenstein]

Entanglement allows system of qubits to have a much larger set of states than combinations of individual qubit states would allow. But there is more to it. It allows for a new behavior where a state of entire system of entangled qubits changes at once, not by changing states of individual qubits.

Pardon me for continuing with silly questions.

If Qbit1 and Qbit2 are entangled. If the state of Qbit1 is known, can we deduct with certainty the state of Qbit2? If yes, then what is the value added of Qbit2?

Analogy to illustrate the above question. I am holding a book in front of a mirror. The image in the mirror is entangled with the real me. Whatever page I show to the mirror, the mirror just returns the mirrored image of the same page. An external observer who wants to know the content of the page, could read the page from the real me or from my "entangled" image in the mirror. That observer will get the exact same information. The entangled image doesn't add any extra information.

I am sure this is a flawed example to illustrate entanglement. Can you please explain why it is flawed. And the mechanism of why entanglement could generate a much larger set of states? More exactly, why does a set of "2 entangled qbits" have more states than a set of "2 non-entangled qbits"?

 

[Hill]

Pardon me for continuing with silly questions.

If Qbit1 and Qbit2 are entangled. If the state of Qbit1 is known, can we deduct with certainty the state of Qbit2? If yes, then what is the value added of Qbit2?

Analogy to illustrate the above question. I am holding a book in front of a mirror. The image in the mirror is entangled with the real me. Whatever page I show to the mirror, the mirror just returns the mirrored image of the same page. An external observer who wants to know the content of the page, could read the page from the real me or from my "entangled" image in the mirror. That observer will get the exact same information. The entangled image doesn't add any extra information.

I am sure this is a flawed example to illustrate entanglement. Can you please explain why it is flawed. And the mechanism of why entanglement could generate a much larger set of states? More exactly, why does a set of "2 entangled qbits" have more states than a set of "2 non-entangled qbits"?

The mirror example does not illustrate quantum entanglement. There are no classical analogs for quantum entanglement. The image in the mirror is not in quantum entanglement with you.

If qubit1 and qubit2 are entangled, then not only "the state of qubit1" is unknown, but qubit1 does not have a definite state. If it does, it is not entangled.

Here is an example to show how new states appear with entanglement. Let's say qubit 1 can be in any superposition of two basis states, $|0_1 \rangle$ and $|1_1 \rangle$, i.e., its general state is $\alpha_1|0_1 \rangle + \beta_1|1_1 \rangle$. Similarly, qubit2's general state is $\alpha_2|0_2 \rangle + \beta_2|1_2 \rangle$. If we take a combination of such states, we can get a state of the two-qubit system in the form $$(\alpha_1|0_1 \rangle + \beta_1|1_1 \rangle)(\alpha_2|0_2 \rangle + \beta_2|1_2 \rangle) = \alpha_1 \alpha_2 |0_1 0_2 \rangle + \alpha_1 \beta_2 |0_1 1_2 \rangle +\beta_1 \alpha_2 |1_1 0_2 \rangle +\beta_1 \beta_2 |1_1 1_2 \rangle$$ These is a non-entangled state.
However, no $\alpha_1, \alpha_2,\beta_1 ,\beta_2$ would make a state, e.g., $$\frac 1 {\sqrt 2} |0_1 0_2 \rangle + \frac 1 {\sqrt 2} |1_1 1_2 \rangle$$ This is an entangled state. It cannot be made from two non-entangled qubits.

 

[PeroK]

Pardon me for continuing with silly questions.

If Qbit1 and Qbit2 are entangled. If the state of Qbit1 is known, can we deduct with certainty the state of Qbit2? If yes, then what is the value added of Qbit2?

If you want to understand quantum computing, you first need to understand some of the basics of quantum mechanics. That's not so much a "silly" question, but a question that comes from having no understanding of QM.

Post #8 gives the mathematical basis of quantum entanglement. It says that the properties of an entangled system of two particles are described by a single wave-function (or single state vector) and not by two separate wave-functions (or two state vectors).

In that sense, Qbit1 does not have its own (pure) state, as it would if it was not part of an entangled system. Hence your question about the pure state of Qbit1 makes no sense.