I've been following the lectures given by Leonard Susskind on Quantum Entanglement. In Lecture 6, he describes a simplified version of the two-slit experiment so that we can use finite dimensional spaces to describe the system (which is all we've learnt so far).(adsbygoogle = window.adsbygoogle || []).push({});

He also uses what he calls the 'Time Evolution Postulate' which states: if |A> -> (evolves to) |A'> and |B> -> |B'>, then any superposition of the two states must evolve linearly as a|A> + b|B> -> a|A'> + b|B'>.

My question is: assuming that both |A> and |B> are both unit vectors, which until this point in the course was necessary for a state to describe a real-world system, shouldn't we now normalise this state for it to make sense?

It does seem to matter, even though Prof. Susskind suggests that it doesn't in this case.

In the lecture |A> represents the state in which the particle goes through slit A, |B> - particle goes through B. Then the particle lands up on one of N 'detectors'.

He represents this by letting |A> and |B> evolve to a superposition of the N detectors ...

|A> -> |A'> = a1|1> + ... + aN|N>

Then with only slit A open: prob(A, m) = <m|A'> = am*am (am* conjugate to am)

|B> -> b1|1> + ... + bN|N>

With only slit B open: prob(B, m) = <m|B'> = bm*bm

Classical result would be prob(A, B, m) = prob(A, m) + prob(B, m) = am*am + bm*bm

But, by the 'Time Evolution Postulate'

|A> + |B> -> (a1 + b1)|1> + ... + (aN + bN)|N>

So quantum result would be prob(A, B, m) = (am + bm)*(am + bm) = (am*am + bm*bm) + (am*bm + ambm*)

It seems to me that normalising |A> + |B> would give a different probability, so surely it does matter.

Any help would be mucch appreciated.

Paul

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# Normalised states

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