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Very well done. Short, yet covers all the necessary details.

Shameless plug, check out my book https://www.amazon.com/dp/0992001021/noBSLA for an in-depth view of the linear algebra background necessary for quantum computing.

If you know linear algebra well, then quantum mechanics and quantum computing is nothing fancy: just an area of applications (See Chapter 9 on QM). Here is an excerpt: https://minireference.com/static/excerpts/noBSguide2LA_previ...

I'm definitely going to check your book out, but in the mean time I was wondering if you could help me zort something out from this article. Author states:

Check these quantum states out as examples, which cannot be broken into a tensor product of two other states, they are unseparable -

Ex.1: 1/sqrt(2)∣00⟩ + 1/sqrt(2)∣11⟩ !=∣ψ1​⟩⊗∣ψ2​⟩

Ex.2: 1/sqrt(2)∣01⟩ − 1/sqrt(2)∣10⟩ !=∣ψ1​⟩⊗∣ψ2​⟩

So, in all previous 2 qubit examples he showed yielded 4 states (00, 01, 10, 11). Is the author saying that some two-qubit systems can be achieved such that not all 4 possible discrete states can participate in superposition? (i.e. in the system of ex. 1, states 10 and 01 are not possible and in ex. 2, states 00 and 11 are not possible?)

The four states |00⟩, |01⟩, |10⟩, |11⟩ form a basis so any two-qubit state can be written as a linear combination of these vectors: ∣ψ⟩ = a|00⟩ +b|01⟩ +c|10⟩ + d|11⟩. If a certain coefficients in the linear combination are zero, e.g., b and c for the 00+11 state, or a and d for the second one, this doesn't have any special meaning.

The classification of states as separable vs entangled refers to the existence of a local description for the two qubits. Remember that |00⟩ is shorthand for |0⟩⊗|0⟩, meaning the state of the two-qubit system when qubit 1 is in state |0⟩ and qubit 2 is in state |0⟩.

Separable states can be written in the form (α|0⟩+β|1⟩)⊗(γ|0⟩+δ|1⟩) = ∣ψ1​⟩⊗∣ψ2​⟩. Note there is a clear local description for the first qubit ∣ψ1​⟩ and and a separate local description of the state ∣ψ2⟩. If Alice prepares her qubit 1 in the state ∣ψ1​⟩ and Bob prepares his qubit 2 in the state ∣ψ2⟩ then the combine description of their two qubits is what's shown above. The state of the combined system is describable as the tensor product of two separate local descriptions.

Entangled states, on the contrary, are states that cannot be described as the tensor product of two local descriptions. Specifically, there exist configurations a,b,c,d for a two-qubit quantum system such that

      a|00⟩ +b|01⟩ +c|10⟩ + d|11⟩   ≠   (α|0⟩+β|1⟩) ⊗ (γ|0⟩+δ|1⟩)
no matter what choice of α,β,γ,δ you make. The phenomenon of entanglement is a quantum-only thing, that can't really be understood via classical analogies, since classical systems can necessarily be described as the combination of two local descriptions. The examples given are two of the four Bell states, see https://en.wikipedia.org/wiki/Bell_state , but there are many more entangled states. Here is a concrete physics example https://en.wikipedia.org/wiki/Singlet_state#Singlets_and_Ent...

Interestingly, many of the quantum computing experiments perform involve the manipulation of entabgled states because they serve as proof that something quantum is going on...

Holy smokes - that is an amazing response! I can't claim that I'm fully grasping the mathematics of entangled vs separable states at this moment, but what you've written seems very clear and I think if I go through it a few more times with the links, I may finally get it after all these years of bewilderment. Thank you!

Hey, love your book. I don't remember how far back I purchased it, but it was back in the days where it wasn't done and we were getting updates pushed via email.

Keep up the good work!

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