# Many questions about qubit (quantum computing)

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1. Jun 13, 2012

### joy13975

Hello! I am new here and I will try my best to clarify my questions and not violate the forum rules.

Just for the record, I am not particularly good at understanding equation-saturated explanations because I am not an expert. However, I did a lot of searching and I still don't understand these questions(call me dumb or whatever). Often I see others asking the same questions but unanswered. Please don't just refer me to some jargon-filled sites :)

::What I think are true (please correct me if I am wrong)::
-A qubit can exhibit only 1 or 0 upon measurement.
-Any disturbance can cause decoherence, which in turn causes wave function collapse and forces a qubit to assume 1 or 0 depending on probability.
-2 or more qubits can be entangled. When one in a pair is measured to have i.e. an up spin the other is known to have a down spin (in most cases) and that the outcome of one is "in some way co-related" to the other.

::What I am very confused with::

1.1 In a Bloch Sphere, how can a quantum state be represented by a dot on the sphere (or inside) when the state might be the spin (which needs to be an axis/direction)?

2.1 A qubit can be in a superposition of 1 and 0. Does it mean a certain value in-between 1 and 0 or simultaneously both 1 and 0 ?

2.2 Why is it that because they can be in superposition, they are capable of parallel computation? All sources simply say that because qubits can be in superposition therefore they can simultaneously compute many things. Without explanation! :X I thought, even if you perform many operations to a qubit, the final measurement show only a single 1 or 0 answer?

3.1 If measuring a qubit forces it to collapse to 1 or 0, what is the point of being able to superpose in-between 1 and 0? The result is still classical?

3.2 If the first measurement can be random (when not aligned with the measuring axis), doesn't that mean a quantum computer can give random answers?

3.3 If "measurement" would force a qubit to collapse to 1 or 0, why wouldn't an "operation"? I mean, both has to introduce external forces and why is it that changing the state of a qubit without looking at it would not force it to collapse? (Or would it? Then doesn't that imply the operation on an entangled qubit would break the entanglement, thus the entanglement is useless in manipulation of qubits?)

3.4 Except for the ion trap method, could you please name one or two other methods of practically measuring the state of a qubit?

4.1 When one in an entangled pair is measured, does the entanglement cease to exist (since the collapse of one instantly collapses the other)? If they disentangle upon measurement, what is the point of measuring the entangled partner instead of just measuring the qubit you want to see directly? (Since both ways collapse them?)

4.2 When two qubits are entangled, does that mean doing operations on one of it affects the other instantaneously in "some co-related way"? Some say it's a misconception.

4.3 Could you name a couple practical methods of doing operations on a qubit? e.g. flip the spin of a photon through using some electromagnetic wave. I just need the names and I will find out by myself.

5.1. Is it true that the faintest environmental disturbance (e.g. a very weak earthquake near a quantum computer, may exert microscopic forces on the qubits) will cause decoherence thus making the results totally nonsense? (This make a quantum computer extremely fragile and prone to break down..?)

I would be grateful if somebody even attempted to answer some of it :) These are bugging me so much.

Please include where you learned the answers from (does not have to be official sources) so I have a rough idea of its credulity, thank you very much!

P.S. This is not a homework. I am just confused.

2. Jun 13, 2012

### fissile_mop

The spin vector is really only useful for calculations done on groups of particles. Also since a qubit is would only contain the spin-up or spin-down information, or a superposition of either, the 'direction' or 'axis' is irrelevant.

It means both 0 and 1 at the same time, with time-varying probabilities that either one will become realized when the wave-function collapses.

I'm not sure as you didn't provide a source to put it into context, but I believe that entanglement provides the means for parallel transfer of information from one machine to another, or from one computational area to another, and not necessarily simultaneous computation on one qubit.

The point is that you have probabilisitic algorithms built into each qubit. In a deterministic computer, let's say a 2-bit, the chance of a piece of information being either 00, 01, 10, or 11 is exactly 1. In a quantum computer, there is a probability that a piece of information will be in any of these states. This allows you to create algorithms on quantum computers which work differently than deterministic computers - some of which are better suited for some calculations that others.

Yes, the calculations returned by quantum computers will have a randomness associated with them. Quantum computers will have to give an answer with a statistical measurement of how accurate it is. In this way, algorithms will need to be repeated in order to increase accuracy to a degree that is acceptable.

I think the architecture and programming of such a computer would be based on the fact that if you set the state of one qubit you will be setting the state of the entangled qubit.

That's all I can focus my attention on for right now. Maybe someone else can address the rest of your questions.

3. Jun 14, 2012

### DrewD

Not exactly, but random disturbances from the environment are a major concern. There are multiple ways to combat this. One is to be sure that the physical qubit is isolated as much as possible to begin with. As a part of this there may be an active control on the qubit that will help keep it isolated. That is very difficult.
Error correcting codes are the best defense. These are used with classical computers too. They work by making any individual error in a set of qubits change the qubits to a different set that has the same meaning. The most basic way of illustrating this is the classical case where you have 0 and 1 as your two options for a piece of info. Instead of encoding them as simply 0 and 1 where an error would completely screw things up you encode 0=000 and 1=111 and then flipping one of the zeros or ones still makes 2/3 of the bits the correct value. The computer can correct for this. It gets more complicated with qubits

Do you mean the physical methods of measuring and changing the value of a qubit or string of qubits? I don't know much about the experimental side. It would depend a lot on what is being used for the qubit. I can't really answer this question, but I do want to note that a qubit, like a classical bit, is not one specific physical thing. It is a piece of information and there are many different ways to store that information. Hopefully somebody else who knows more about experimental apparatus will be able to answer this.

If you have two entangled qubits, knowing one determines the other, but only if you knew the initial state. The example used to talk about EPR is a zero total spin pair of 1/2 spin particles. If one is +1/2, the other is -1/2, but that only true IF you knew that they add to 0. So it may not be as easy as measuring just one. To answer the second part, I know of no experimental evidence can show that the effect on the combined state is not instantaneous. The math seems to support an instantaneous change and that's good enough for me.

Why would it not be aligned? A qubit will be prepared in a certain way and measured in a way that would have been decided knowing the options for preparation. Often, a computation qubit will be more than a single particle. This allows for a large space of possible measurement outcomes. Many of those outcomes will have the same computational meaning. I'm going to disagree with fissile_mop on this unless I misunderstood the question. I maybe wrong, but my understanding is that the quantum algorithms are designed specifically to work within (and utilize) the randomness of a quantum measurement.

I just realized that you may mean, "can one purposefully get a random answer from a quantum computer?" Yes! There could be a new fundamental physical basis for rand()!!!

4. Jun 14, 2012

### chogg

Welcome!

The axis goes from the center of the sphere to the point on the sphere. It's just a convenient way to encode the axis.

5. Jun 14, 2012

### Delta Kilo

I'll give it a go (disclaimer: I'm not exactly an expert by I try to be careful ant not to post nonsence)

Yes.
Yes. The disturbance has to be by a macroscopic object (with large number of coupled degrees of freedom) and it has to leave a trace in that object.
Yes. But it is al lot more interesting in case of 3 or more qubits. For 1 qubit you have 2 orthogonal state vectors plus all their linear combinations. For N qubits you get 2N orthogonal vectors, in other words a separate 'axis' in state space for each of the 2N numbers which can be represented by N bits.
For spin 1/2 particle (electron), the radius vector of the point on a Bloch sphere corresponds to the direction of the particle's spin in 3D space BUT it is a lucky coincidence (due to SU(2) being double cover for SO(3)). For photon (spin-1 particle), points at the top and bottom of Bloch sphere may correspond to vertical and horizontal polarizations, points at left and right to 45/135 degrees diagonal and front and back to left and right circular polarization.
The important thing is, spin lives in its own spin space, not in the usual euclidian 3D. In this space you have 2 orthogonal axis defined by spin states |Z+> and |Z-> (that is pointing up and down) and spin in any other direction is represented as a linear combination of those with complex coefficients:
|x+> = 1/√2 (|z+> + |z->),
|x-> = 1/√2 (|z+> - |z->) ,
|y+> = 1/√2 (|z+> + i|z->),
|y-> = 1/√2 (|z+> - i|z->)
etc. and any point on the Bloch sphere can be represented this way. The choice of |z+>,|z-> is arbitrary, any pair of orthogonal states can be used as a basis.
It is important to understand that states |0> and |1> are not just marks 0 and 1 along some axis, but rather two orthogonal unit vectors and any other state, being a linear combination of those, will have some |0>-component and some |1>-component.

Because Shroedinger evolution is linear, any combination of two or more solutions is also a solution. Suppose you have a quantum logic circuit with N inputs and N outputs, computing a function y=f(x). If you feed it a superposition of all possible inputs you get back a superposition of all answers.
Yes. You can only get N real bits out of N qubits. But that is often sufficient. For example, in Grover's search algorithm, you want to find x such that some f(x)=1. So you create a circuit which computes f(x) and then inverts the phase of x if f(x)=1 and leave it intact otherwise. Then you feed it a superposition of all possible x's and get back the same superposition where the correct x's have their phase inverted. With a bit of fiddling you can arrange it so that when you do a measurement you get the correct x with high probability.
Depends on the algorithm. There are both kinds.
Depends on whether the interaction leaves any traces of it in the external environment, i.e. whether the information is allowed to leak out. (rough analogy is conservative vs. dissipative forces in classical mechanics)
General state of two entangled qubits is |X> = a|00> + b|01> + c|10> + d|11>. When you measure the first qubit you get 0 with probability p(0) = |a|^2+|b|^2 in which case the state of second qubit becomes a|0>+b|1> (multiplied by appropriate normalization constant) , or you get 1 with probability p(1)=|c|^2+|d|^2 and the state of the second qubit becomes c|0>+d|1>. So the second qubit may still carry some useful info. Often when people talk about strongly entangled qubits they mean either |00>+|11> or |01>+|10>, in which case yes, the second measurement is redundant.
Hard to tell exactly when and how this happens, but when you compare the results afterwards you see the coincidences. And, according to Bell's theorem, there are more coincidences that can possibly be explained in "local realistic" way.

6. Jun 24, 2012

### joy13975

The answers you have provided are very helpful.

Thank you very much guys!

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