3 ms·

I'm a little worried about something in quantum computing. If I treat my quantum computer as just a channel for sending a message to myself I think I can invok

by ouid 7y ago

I'm a little worried about something in quantum computing. If I treat my quantum computer as just a channel for sending a message to myself I think I can invoke the channel capacity theorem. Furthermore, I think that if I try to read an entangled state, it's as if I am reading the entire state all at once, so the maximum number of bits I should be able to read is bounded by the logarithm of the ratio of the signal to the noise, where noise scales with temperature. That number doesn't scale well. It suggests that if I want to read an additional bit from a quantum computation, I need to drop the temperature by half. Furthermore, I introduce heat to my quantum computer when I input the state. I must construct the input to my quantum computer from a uniform mixture by the no cloning principle. This is going to generate heat, and the rate at which I can dissipate that heat is bounded by the speed of light, This gives me an upper bound on my temperature of O(1/t^3) assuming extremely optimal heat dissipation. This means that the time required to cool my state to the point where I can read n entangled bits is exponential in n, right?

comnetxr 7y ago
No. I think this is where your logic is off: "Furthermore, I think that if I try to read an entangled state, it's as if I am reading the entire state all at once" You can't read the entire state. A quantum channel where you send n-bits lets you read n-bits, not exponential in n-bits. There is indeed reduction of signal as you increase temperature. Under the error correction theorem, you can amplify that signal back to full strength as long as you are under a threshold temperature (but with ever-increasing resources as you approach the threshold temperature.) Also this: "This is going to generate heat, and the rate at which I can dissipate that heat is bounded by the speed of light". Sure, the heat has to be transferred out of the system, which takes time that is bounded by some distance divided by the speed of light. But the qubit is on a 2D chip surrounded by a 3D refrigerator. The distance doesn't necessarily increase with increasing qubits.
- ouid 7y ago
  >You can't read the entire state. A quantum channel where you send n-bits lets you read n-bits, not exponential in n-bits. An entangled state is different, Since I can read all of the bits outside of each other's light cones, they cannot causally influence each other, which is why I said as if instead. There is no such thing as "all at once", but I can properly read each qubit before all of the others, as far as that qubit is concerned and it would still obey the correlation. The reading of an entangled quantum state is indeed a single measurement. >Under the error correction theorem, you can amplify that signal back to full strength as long as you are under a threshold temperature (but with ever-increasing resources as you approach the threshold temperature.) 2) I don't think that quantum error correction ultimately solves the problem of heat. It tries to solve the problem of single bit/sign flips from not doing your computation in a closed system. My argument assumes that the only things that exist in the entire universe are the computer and the heat. "This is going to generate heat, and the rate at which I can dissipate that heat is bounded by the speed of light". Sure, the heat has to be transferred out of the system, which takes time that is bounded by some distance divided by the speed of light. But the qubit is on a 2D chip surrounded by a 3D refrigerator. The distance doesn't necessarily increase with increasing qubits. 3) you're misinterpreting my statement of the problem. I don't have to get each additional qubit just as cold as the original. Every qubit I add requires me to get the entire quantum state to half the temperature that I had previously. The geometry of your refrigerator here only matters in that it's finite dimensional.

by ouid 7y ago

comnetxr 7y ago

No. I think this is where your logic is off: "Furthermore, I think that if I try to read an entangled state, it's as if I am reading the entire state all at once" You can't read the entire state. A quantum channel where you send n-bits lets you read n-bits, not exponential in n-bits. There is indeed reduction of signal as you increase temperature. Under the error correction theorem, you can amplify that signal back to full strength as long as you are under a threshold temperature (but with ever-increasing resources as you approach the threshold temperature.) Also this: "This is going to generate heat, and the rate at which I can dissipate that heat is bounded by the speed of light". Sure, the heat has to be transferred out of the system, which takes time that is bounded by some distance divided by the speed of light. But the qubit is on a 2D chip surrounded by a 3D refrigerator. The distance doesn't necessarily increase with increasing qubits.

ouid 7y ago
>You can't read the entire state. A quantum channel where you send n-bits lets you read n-bits, not exponential in n-bits. An entangled state is different, Since I can read all of the bits outside of each other's light cones, they cannot causally influence each other, which is why I said as if instead. There is no such thing as "all at once", but I can properly read each qubit before all of the others, as far as that qubit is concerned and it would still obey the correlation. The reading of an entangled quantum state is indeed a single measurement. >Under the error correction theorem, you can amplify that signal back to full strength as long as you are under a threshold temperature (but with ever-increasing resources as you approach the threshold temperature.) 2) I don't think that quantum error correction ultimately solves the problem of heat. It tries to solve the problem of single bit/sign flips from not doing your computation in a closed system. My argument assumes that the only things that exist in the entire universe are the computer and the heat. "This is going to generate heat, and the rate at which I can dissipate that heat is bounded by the speed of light". Sure, the heat has to be transferred out of the system, which takes time that is bounded by some distance divided by the speed of light. But the qubit is on a 2D chip surrounded by a 3D refrigerator. The distance doesn't necessarily increase with increasing qubits. 3) you're misinterpreting my statement of the problem. I don't have to get each additional qubit just as cold as the original. Every qubit I add requires me to get the entire quantum state to half the temperature that I had previously. The geometry of your refrigerator here only matters in that it's finite dimensional.