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Microsoft told me that any operation a quantum computer performs HAS to be reversible: something like XOR doesn't work because there're four possible inputs and

Microsoft told me that any operation a quantum computer performs HAS to be reversible: something like XOR doesn't work because there're four possible inputs and only two possible outputs, so the state of the inputs is destroyed. However, something like NOT works fine because the input-output mapping is 1:1. It seems like cheating to say that "reversing cause and effect is no trouble" when the computer was never allowed to perform irreversible operations to begin with. I think I am missing some important information or maybe I am misunderstanding the words used in the article?

emiraga 8y ago
You have the right idea, but not the right operation. XOR is reversible. To give a better example: COPY is not reversible. Imagine that you have two qubits [A, B] as an input and you want to output [A, A]. That is not possible with quantum computers.
- mistaken 8y ago
  Isn't everything reversible if enough information is stored? Instead of COPY [A, B] -> [A, A]; you could have [A, B, 0] -> [A, A, B] which saves the value of B.
  - ahakki 8y ago
    IIRC this is how you implement stuff, you always keep the inputs, so everything stays reversible.
    - lisper 8y ago
      Yes, but "keeping the inputs" is non-trivial in quantum-land because of the no-cloning theorem. You can't just make a copy of your inputs and stash them. You have to be more clever than that.
  - deleted 8y ago
    [deleted]
- Aardwolf 8y ago
  XOR has 2 inputs and 1 output so cannot be reversible. CNOT is the closest reversible gate
  - im3w1l 8y ago
    To elaborate: CNOT outputs the pair (A xor B, B). By also outputting B, it becomes reversible.
toxicFork 8y ago
Note: I'm not a professional quantum scientist so if any real ones come along please tell me if I'm getting this right, and please correct any mistakes that you see in my logic. From my understanding so far: One benefit of using qubits is that you can map multiple of them to one traditional bit. Let's say we use light for our quantum computer. I like to see a wave of light as a continuous stream of photons, and photons as "an order". I'm assuming all waves and photons in the rest of this comment are of the same frequency. If you have a vertically polarised wave of light, it orders you this: "at time 1: jump up. at time 2: jump down. at time 3: jump up, ..." If you have a horizontally polarised photon, it orders you this: "at time 1: move left. at time 2: move right. at time 3: move left, ..." We can start by saying "horizontal polarization is 0" and "vertical polarization is 1". H = 0 V = 1 If we define XOR using only H and V: H ^ H = H (0) H ^ V = V (1) V ^ H = V (1) V ^ V = H (0) Let's now add phase to the qubits to enhance them. Phase can be seen as adding a time delay (or "shift") to all of the photons. Let's see the vertically polarized photon again: "at time 1: jump up. at time 2: jump down. at time 3: jump up, ..." And at 180 degree phase, you shift all orders by 1 time unit: "at time 1: jump down. at time 2: jump up. at time 3: jump down, ..." One fun and useful property of phase is that if you have two photons of the same polarization but they are at phases 180 degrees apart from each other, they will cancel each other out because you cannot move up and down at the same time. Getting back to qubits and XOR: Let's say H at 180 phase is I, and V at 180 phase is W. Note that I is still horizontally polarized, and W is still vertically polarized. Let's bring them together, where we have 4 types of qubits mapped to 2 types of bits: H = 0 I = 0 V = 1 W = 1 Let's add a step to the XOR operation: - if the right input is V, phase shift the output by 180 degrees The outputs of our "XOR 2.0: quantum edition": H ^ H = H, no shift = H (0) H ^ V = V, shift it = W (1) V ^ H = V, no shift = V (1) V ^ V = H, shift it = I (0) Now we have output that still matches the original definition of XOR when mapped to traditional bits, but with the bonus of reversibility. H => 0,0 W => 0,1 V => 1,0 I => 1,1
ryanr1230 8y ago
This is true but you can use extra qubits that when ignored make the output behave like XOR, so its not as limiting as you might think

emiraga 8y ago

You have the right idea, but not the right operation. XOR is reversible. To give a better example: COPY is not reversible. Imagine that you have two qubits [A, B] as an input and you want to output [A, A]. That is not possible with quantum computers.

mistaken 8y ago
Isn't everything reversible if enough information is stored? Instead of COPY [A, B] -> [A, A]; you could have [A, B, 0] -> [A, A, B] which saves the value of B.
- ahakki 8y ago
  IIRC this is how you implement stuff, you always keep the inputs, so everything stays reversible.
  - lisper 8y ago
    Yes, but "keeping the inputs" is non-trivial in quantum-land because of the no-cloning theorem. You can't just make a copy of your inputs and stash them. You have to be more clever than that.
- deleted 8y ago
  [deleted]
Aardwolf 8y ago
XOR has 2 inputs and 1 output so cannot be reversible. CNOT is the closest reversible gate
- im3w1l 8y ago
  To elaborate: CNOT outputs the pair (A xor B, B). By also outputting B, it becomes reversible.

toxicFork 8y ago

Note: I'm not a professional quantum scientist so if any real ones come along please tell me if I'm getting this right, and please correct any mistakes that you see in my logic. From my understanding so far: One benefit of using qubits is that you can map multiple of them to one traditional bit. Let's say we use light for our quantum computer. I like to see a wave of light as a continuous stream of photons, and photons as "an order". I'm assuming all waves and photons in the rest of this comment are of the same frequency. If you have a vertically polarised wave of light, it orders you this: "at time 1: jump up. at time 2: jump down. at time 3: jump up, ..." If you have a horizontally polarised photon, it orders you this: "at time 1: move left. at time 2: move right. at time 3: move left, ..." We can start by saying "horizontal polarization is 0" and "vertical polarization is 1". H = 0 V = 1 If we define XOR using only H and V: H ^ H = H (0) H ^ V = V (1) V ^ H = V (1) V ^ V = H (0) Let's now add phase to the qubits to enhance them. Phase can be seen as adding a time delay (or "shift") to all of the photons. Let's see the vertically polarized photon again: "at time 1: jump up. at time 2: jump down. at time 3: jump up, ..." And at 180 degree phase, you shift all orders by 1 time unit: "at time 1: jump down. at time 2: jump up. at time 3: jump down, ..." One fun and useful property of phase is that if you have two photons of the same polarization but they are at phases 180 degrees apart from each other, they will cancel each other out because you cannot move up and down at the same time. Getting back to qubits and XOR: Let's say H at 180 phase is I, and V at 180 phase is W. Note that I is still horizontally polarized, and W is still vertically polarized. Let's bring them together, where we have 4 types of qubits mapped to 2 types of bits: H = 0 I = 0 V = 1 W = 1 Let's add a step to the XOR operation: - if the right input is V, phase shift the output by 180 degrees The outputs of our "XOR 2.0: quantum edition": H ^ H = H, no shift = H (0) H ^ V = V, shift it = W (1) V ^ H = V, no shift = V (1) V ^ V = H, shift it = I (0) Now we have output that still matches the original definition of XOR when mapped to traditional bits, but with the bonus of reversibility. H => 0,0 W => 0,1 V => 1,0 I => 1,1

ryanr1230 8y ago

This is true but you can use extra qubits that when ignored make the output behave like XOR, so its not as limiting as you might think