您的位置:站长主页 -> 繁星客栈 -> 网友原创作品集 -> sage 文集 December 22, 2024

Measurement and Unitarity in QM

- sage -

Here is a rather long and somewhat technical note of it. I apologize for not giving a very detailed and down to earth explanation.

Sorry about having to introduce density matrix. But I find it is the most closest analogy to the Liouville theorem,
-----------------------------------------------------------------------------
-----------------------------------------------------------------------------
Roughly speaking, unitarity means conservation of probability.
Suppose I start with a state |psi>. After a while, the state evolves
to U|psi>. Then, conservation of probability means that
<psi|U^+ U |psi>=<psi|psi>. It requires the evolution matrix to be
unitary, i.e., U^+ U=1.

More strictly speaking, unitarity is somewhat like the Liouville
theorem. It is the conservation of some probability 'flow'...

It is the one of the biggest mysteries that measurement irreversibly
destroy this flow. This can be illustrated just by the double-slit
experiment.

----------------------------------------------------
Before doing that, just to be systematic, let me introduce a little
bit of language, the density matrix. For any single quantum mechanical state
|psi>, define density matrix by

W=|psi><psi|.

First, it is a matrix. Given any base |1>, |2>....|n>....., matrix
element W_mn = <m|psi><psi|n>. It is called density matrix since the
expectation value of any operator O is

<O>=Tr (W O)

where the trace is carried out over any complete basis. This is again
easy to see as

<O> =<psi|O|psi>= Sum_n,m <psi|m><m|O|n><n|psi>

= Sum <n| ( |psi><psi|) O |n> = Trace (W O).

Therefore, if you like, density matrix is a statistical weight (which
is analogous to the quantity in Liouville Theorem). We can also show
that Tr(W)=Sum <n|psi><psi|n> = Sum <psi|n><n|psi>=
<psi|psi>=1. Moreover, we can show that Tr(W W )=Tr(W W W....)=1



----------------------------------------------------------------

Now the first important result, a sort of 'quantum Liouville'
theorem:

Any unitary evolution will 'conserve' density matrix of a single
quantum mechanical state.

Let's understand this more carefully. After evolution U, the density
matrix will be

W -> U|psi><psi|U^+

Now it is easy to show Tr(W)=1 is not affected by the evolution since

Tr (U|psi><psi|U^+ ) = Tr (U^+ U|psi><psi| ) = Tr ( |psi><psi|)=1.

Moreover, Tr(W W) should also not to be affected by any unitary
evolution.

...

This is what we mean by 'conserve'.

-----------------------------------------------------------------

Now we come back to double-slit experiment. Say there are slit 1 and
2. The incoming particle is in a state 1/sqrt(2) *(|1>+|2>). Then if
we put a screen after the slit, we see interference pattern...

The density matrix for the incoming state is

1/2(|1>+|2>)(<1|+<2|),

or, in a matrix form in the base of |1> and |2>

| 1/2 1/2 |
| | .
| 1/2 1/2 |

one can easily verify that this staisfies Tr(W)=1,
Tr|W^2|=1,.... Notice that the off-diagonal entry in the density
matrix is very important. It is the memory of a state which could be
both 1 and 2, both here and there. It related to all quantum
mechanical magic/mystery/misery/misunderstanding about coherence, pure
state, etc. It is crucial to make Tr(W^2)=1.

Now suppose we really want to find out what is going on. In
particular, we put detectors at each slit to find out which slit the
particle does pass through. Quantum mechanics rule (it is the RULE,
no explanation ) says that too bad,
now we have made the damn measurement. We destroyed the coherence. The
wave-packet of every partile passing through collapses. The
result would be EITHER |1> OR |2> (not BOTH 1 AND 2). We all know that
it means no interference, and so on.

Suppose we still want to construct a density matrix for the system
after that measurement, it would have to be

| 1/2 0 |
| | .
| 0 1/2 |

since there would be no interference anymore. Immediately, we see that
Tr(W^2)=1/2 not 1!!

We have proved that any unitary evolution cannot change
Tr|W^2|. Therefore, something which is not unitary must have happened
during the measurement.


>我记得,有人认为要把环境和测量仪器(包括观测者)考虑在内作为一个更大的
>系统来考察,而不是象传统那样只分析被观测者,只是不知道这种考虑管不管用。

This might be called decoherence propelled by Wheeler company. I don't
know that much about it. Thinking naively, if all matter obey
schrodinger equation (or whatever Hermitian generator), then the
evolution will always be unitary....

including enviroment means that the dimension of the density matrix is
much larger. In principle, the off-diagonal entries could move to
other places. therefore, if we just focus on the 2 by 2 submatrix, it
is possible that it is diagonal. However, the coherence is somewhere
else. This does not seem solve the basic measurement problem
especially if we are allowed to think of the universe as a big pure
state... This would force us to think of some multiworld theory or
something like that. maybe we are not allowed to say universe a pure
state since gravity is very mysterious. but this is certainly just
speculation.  

anyway, I hope this will tell us something new but I don't see even a
vague hint of it. I am still too young to work on that...

二零零五年六月十五日