Quantum computations, irreversibility, unitarity

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SUMMARY

In quantum computations, operators applied to initial quantum states must be unitary, ensuring reversibility. This is crucial for maintaining the physical integrity of quantum setups. Deutsch's algorithm exemplifies this by distinguishing between balanced and constant functions using a single quantum operation on two qubits. However, the function f(x) itself cannot be represented by a quantum operator if it is constant, as it is not reversible and thus cannot correspond to a unitary operator.

PREREQUISITES
  • Understanding of quantum states and operators
  • Familiarity with unitary operations in quantum mechanics
  • Knowledge of Deutsch's algorithm and its application
  • Concept of balanced and constant functions in quantum computing
NEXT STEPS
  • Study the principles of unitary operators in quantum mechanics
  • Explore the implementation of Deutsch's algorithm in quantum programming languages
  • Learn about reversible and irreversible processes in quantum computations
  • Investigate the implications of non-reversible functions in quantum algorithms
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Quantum computing enthusiasts, researchers in quantum algorithms, and students studying quantum mechanics who seek to deepen their understanding of unitary operations and their applications in algorithms like Deutsch's.

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A stupid couple of questions...

In quantum computations, one typically starts with some initial quantum state on which an operator is applied.

This operator must be unitary , right? (I guess that otherwise, it would not corerspond to an actual physical quantum setup). And this implies that the operator must be reversible, correct?
So in quantum computing, we consider only irreversible processes, correct?


Now consider Deutsch's algorithm. It deals with a function acting on a single qubit which may be either "balanced" or "constant".

It is balanced if it may give both 0 and 1 a sresult. One possibility is f(0)=0, f(1)=1 and the other balanced possibility is f(0)=1, f(1)=0.

It is constant if it gives the samr answer no matter what the input is. So the first possibility is f(0) =f(1) =0 and the second possibility is f(0)=f(1)=1.

Now, the whole idea of Deutsch's algorithm is to able to distinguish whether f is balanced or constant with a single application of a certain operation which must be chosen judiciously. (and the input state must be chosen judiciously as well).

Now, usually people say: classically, one cannot determine if the function is constant or balanced without two runs fo the function. (for example, observing that f(0) gives zero does not tell us if it's balanced or not). Then people say that quantum mechanically, one can tell if f is balanced or not with a single run of an operator applied to a suitable combination of two qubits.

Fine. But my question is this: if I understand correctly, the function f(x) itself cannot be represented by any quantum operator , correct? Because in the constant case it is not reversible and therefore could not be associated to a unitary operator.

This kind of surprised me when I thought about this today...It's weird in some sense because the starting point is the operator "f" and yet I have never seen anyone mentioning that f itself cannot be implemented as a quantum operation. Maybe I am missing something?
 
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He probably means something like the following operation on two qubits:

|x>|y> ---> |x>|y + f(x)>
 
Hurkyl said:
He probably means something like the following operation on two qubits:

|x>|y> ---> |x>|y + f(x)>

Yes, I know this is the operation used in Deutsch's algorithm. I was wondering if I was correct in saying that the operation on a single qubit |x> ---> |f(x)> would be impossible to implement.
 

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