Why must the value of a qubit in quantum computing remain secret?

Question

I've read time and again, as I dig through all the available literature, that in quantum computing, the smallest unit of value--a qubit--must remain "secret" or unknown until such a time as it is measured. In StackOverflow, I even read that, "in order for a qubit to work as one, its state has to be secret from the rest of the physical universe, not just from you. It has to be secret from wisps of air, from nearby atoms, etc. On the other hand, for qubits to be useful for a quantum computer, there has to be a way to manipulate them while keeping their state a secret. Otherwise its quantum randomness or quantum coherence is wrecked" (source: Does anyone know what "Quantum Computing" is?, answered by Greg Kuperberg). This notion of the the secrecy of a qubit went beyond anything I've read so far, but nonetheless, why should this be...I mean, what explains and justifies this strange property--this secrecy or unmeasurability of a qubit? Hopefully the answer to this question will help me to start making the mental transition from classical computing machines.

Answer 1

It's because quantum stuff only interferes if every detail everywhere ends up the same.

For example, the Hadamard operation H sends the state |0⟩ to √½|0⟩+√½|1⟩ and the state |1⟩ to √½|0⟩-√½|1⟩.

H |0⟩ = √½|0⟩ + √½|1⟩
H |1⟩ = √½|0⟩ - √½|1⟩

A neat thing about H is that it is its own inverse: if you apply it twice, it undoes itself.

H H |0⟩ = H (√½|0⟩ + √½|1⟩)
        = √½ H |0⟩ + √½ H |1⟩
        = √½ (√½|0⟩ + √½|1⟩) + √½ (√½|0⟩ - √½|1⟩)
        = ½ |0⟩ + ½|1⟩ + ½|0⟩ - ½|1⟩
        = (½+½) |0⟩ + (½-½) |1⟩
        = |0⟩

H H |1⟩ = H (√½|0⟩ - √½|1⟩)
        = √½ H |0⟩ - √½ H |1⟩
        = √½ (√½|0⟩ + √½|1⟩) - √½ (√½|0⟩ - √½|1⟩)
        = ½ |0⟩ + ½|1⟩ - ½|0⟩ + ½|1⟩
        = (½-½) |0⟩ + (½+½) |1⟩
        = |1⟩

But now consider what happens if, between those two Hadamards, we use a controlled-not to attempt to copy the qubit-being-Hadamarded's value onto a second qubit.

Even though we only use the qubit as a control, the own-inverse property breaks:

H₁ C₁NOT₂ H₁ |00⟩ = H₁ C₁NOT₂ H₁ |0⟩⊗|0⟩
                  = H₁ C₁NOT₂ (H|0⟩)⊗|0⟩
                  = H₁ C₁NOT₂ (√½|0⟩ + √½|1⟩)⊗|0⟩
                  = H₁ C₁NOT₂ (√½|00⟩ + √½|10⟩)
                  = H₁ (√½|00⟩ + √½|11⟩)
                  = √½ H₁ |00⟩ + √½ H₁ |11⟩
                  = √½ (H|0⟩)⊗|0⟩ + √½ (H|1⟩)⊗|1⟩
                  = √½ (√½|0⟩ + √½|1⟩)⊗|0⟩ + √½ H (√½|0⟩ - √½|1⟩)⊗|1⟩
                  = ½|00⟩ + ½|10⟩ + ½|01⟩ - ½|11⟩

The second qubit adds more room to the state space, and the CNOT moves some of our state into that extra room. So instead of the computation folding the states back in on themselves to cause destructively interference they... just kinda spread out.

Without destructive interference, you might as well just be flipping coins instead of rotating qubits. So carefully managing this effect is very important in quantum computation.

You can try the example for yourself in the toy circuit simulator Quirk, which has inline state displays:

Answer 2

Since you quoted my own answer to another SO question, I hope that I can give you something of a conceptual answer. It's one of the principles of quantum probability that if you measure a property of a quantum object, you always might change its state. This is illustrated for instance in the Stern-Gerlach experiment with electron spin, which is described very nicely in the Feynman Lectures on Physics. The spin state of an electron is a clean example of a qubit and it is very convenient for thought experiments (even though it is not popular at this time for qubit implementations in QC technologies). You can measure whether a qubit is spin UP or spin DOWN, or you can measure whether it is spin LEFT or spin RIGHT. If you measure the spin in the same direction twice in a row, you will get the same answer, so that the qubit can (among other things) act like an ordinary bit. However, if the qubit is spin RIGHT, and if you then measure its spin vertically, then that measurement has the effect of erasing the answer to the horizontal spin measurement. I.e., you will get an answer of either UP or DOWN, and for either answer, a horizontal spin measurement afterwards will be split 50-50 between UP and DOWN.

This is just one example of a more general principle, that two measurements can interfere with each other. (Mathematically, the measurements might not commute.) Moreover, what matters is not whether you personally carry out the measurements, but rather whether any entity measures your qubit, or in other words whether any entity interacts with its qubit state. These delicate probabilities, that can be ruined by non-commuting measurements, are exactly what powers quantum computation as, what I called it before, "randomized computation on steroids". Thus the qubits must be kept secret until the end of the computation, or else the rules of quantum probability will be wrecked and the quantum computer will degenerate to (at best) a classical computer with access to ordinary randomness.

In this answer I'm not saying much about what exactly is different about quantum probability. Well, that is not an easy topic and if you want to learn it, I would recommend a textbook such as Nielsen and Chuang. But part of the essence of it is that in quantum probability, different probabilistic histories can "interfere". This is illustrated for instance in the two-slit experiment, where a photon has some probability of making it through either of two slits to reach a detector. But if both slits are open, the probability (or more precisely the quantum amplitudes that yield the probabilities) can cancel; or they can reinforce each other to produce an amplified probability that is greater than the chance of passing through either slit alone. Precisely because these effects violate the normal rules of probability, they require secrecy, i.e., the effect is wrecked if any entity witnesses which slit the photon travelled through.