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In quantum cryptography why do we need the Heisenberg uncertainty principle?


I only know the statement of the Heisenberg uncertainity principle. As I know that if Eve tries to know the polarization angle of a photon with a wrong basis then it will alert others. Then where is the use of Heisenberg uncertainity rule in it? I know this is a childish question to physics people but I really need to know the answer. Please anybody answer it.

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Welcome to Physics.SE. There is probably a good question in here, but right now it is too underspecified to be reasonably answerable. Can you make this more specific in terms of (1) you understanding of the Heisenberg rule in general and (2) why you think it might not be important in any quantum process? – dmckee Mar 28 '12 at 14:21
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Heisenberg's uncertainty principle in its most general form is a statement about the measurement uncertainty (or variance) of so-called non-commuting variables.

In quantum mechanics, everything you can observe (such as the polarization of a photon) is represented by a so-called operator. By performing a measurement, you act with an operator on a quantum state.

Two operators $A$ and $B$ are said to commute if $AB = BA$. In simple terms, for commuting operators it doesn't matter whether you first measure $A$ and then $B$ or first measure $B$ and then $A$.

If, however, the operators do not commute, then the order matters: If you measure first $A$ and then $B$, you get a different outcome than if you measure first $B$ and then $A$.

One example is position and momentum: If you measure the position first, then the momentum will have a very high uncertainty. If you measure momentum first, then position will have a very high uncertainty.

The relevance for quantum cryptograhpy now is that measuring the polarization angle in a basis $1$ and measuring it in a basis $2$ corresponds to two non-commuting operators. This means that Eve measuring with basis $2$ before Bob measures with basis $1$ will lead to different outcomes than if Eve would do nothing. Heisenberg basically says that if Eve measures in basis $2$, then the uncertainty of the observable "Polarization in basis 2" is zero and therefore the uncertainty of the observable "Polarization in basis 1" must be at a maximum, i.e., Eve has completely messed up the state of the photon for basis $1$.

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