# Non disturbing (?) measurement

## A or B ?

2 vote(s)
100.0%
2. ### B

0 vote(s)
0.0%
1. Nov 22, 2005

### kleinwolf

Let's take a spin 1/2 system, and the trivial operator : $$A=\mathbb{I}=\left(\begin{array}{cc} 1 & 0\\0 & 1\end{array}\right)$$.
Suppose X=(1,0) is the initial state before the measurement. Question :
A) Is the final state Y=X, for obvious reasons.
B)The eigenspace for the eigenvalue 1 of A is the whole R^2. A particular normalized eigenvector is parametrized by a polar angle and given by : $$Y=(\cos(\phi),\sin(\phi))$$. Y is the endstate of this particular measurement, and $$p(Endstate=Y)=|\langle X|Y\rangle|^2=\cos(\phi)^2$$.
Is A or B the correct answer ?
Remarks :
1) A is a particular case of B
2) How do you interprete $$\phi$$ physically in the context of quantum-mechanics ?

2. Nov 22, 2005

### vanesch

Staff Emeritus
An ideal measurement gives you BY DEFINITION, A. Otherwise, the measurement is called non-ideal. A non-ideal measurement is not entirely determined by its hermitean operator if there is degeneracy. A non-ideal measurement can be seen as an ideal measurement followed by a unitary operator ; it is necessary to specify the unitary operator (which depends on the precise experimental setup). In your case, it is determined by phi.

3. Nov 22, 2005

### kleinwolf

So if I have the operator A=diag(1,1,-1)...I measure x=1/sqrt(2)*(1,0,1)...so that if the result is -1 the measurement is forcedly ideal, where as for 1 not forcedly...so is generally speaking (with taking into account prob. of outcomes) the measurement ideal or not ?

4. Nov 22, 2005

### vanesch

Staff Emeritus
It really depends upon the measurement setup. For instance, as I'm an MWI-er, I don't believe that there is such a thing as collapse, and that the entire unitary evolution corresponding to the physical interactions in the measurement apparatus will determine "pointer states" that correspond to the eigenvectors of the hermitean operator that we associate with a measurement apparatus. As such, it is then rather clear, from that unitary evolution (which is determined by the physics of the measurement apparatus) whether the initial system states get through "unharmed" or whether they get some unitary rotation - in other words, whether my apparatus corresponds to an ideal measurement or not.
But of course in the case that there is only ONE initial state corresponding to a pointer state (non-degenerate case) there's not much choice for a unitary evolution, so in that case the apparatus will be ideal in all cases.

5. Nov 22, 2005

### kleinwolf

the problem with this, is that with the operator A=diag(1,1,-1) and x=1/sqrt(2)(1,0,1) is that there are two kind of unitary operations :

-1) it "collapses" (I know you don't like that word) to the eigenvector corresp. to -1, but the measurement still is ideal

+1)

a) it collapses to y=(1,0,0) and is ideal (at least I suppose from your previous messages)

b) it collapses to y=(cos(phi),sin(phi),0) and there exists phi for which it is not ideal...

How do you make the difference between the two unitary operations that determine if the measurement is ideal or not ?

6. Nov 22, 2005

### vanesch

Staff Emeritus

Well, consider the three state system of your example, with states |1>,
|2> and |3> (so (1,0,0) ; (0,1,0) and (0,0,1) in the basis you use).

Let us consider the two pointer states |psi1> and |psi2> corresponding to the two outcomes of the measurement. Let us call |psi0> the state of the measurement apparatus before the measurement.

If my measurement apparatus corresponds to U1, with U1:

|psi0> |1> ==> |psi1> |1>
|psi0> |2> ==> |psi1> |2>
|psi0> |3> ==> |psi2> |3>

then the measurement apparatus will give you an ideal measurement.

However, the physics of my measurement apparatus could be different, and it could correspond to an operator U2:

|psi0> |1> ==> |psi1> (cos th |1> + sin th |2>)
|psi0> |2> ==> |psi1> (-sin th |1> + cos th |2>)
|psi0> |3> ==> |psi2> |3>

That would be your other case. In both cases, the measurement apparatus corresponds to the hermitean operator you cited, but the second kind of apparatus is not an "ideal measurement" apparatus.

The reason is that the hermitean operator determines not uniquely the unitary evolution of the measurement apparatus, unless you ALSO specify that it is "ideal". The reason is that the hermitean operator is made up of 2 things:
A) The projection operators on the eigenspaces that correspond to the POINTER states ; in this case:
space 1 corresponds to the pointer state |psi1> which is reached from states |1> and |2>, so the projector projects upon the space spanned by |1> and |2> ==> P1

space 2 corresponds to the pointer state |psi2> which is reached from state |3>, so the projector P2 projects onto the space generated by P2.

B) the "measurement values" we assign to these pointer states (the values on the display), say m1 and m2.

The hermitean operator of the measurement is then simply:

m1 P1 + m2 P2

Clearly, what happens to the state of the system is not included in that description, and both U1 and U2 give rise to the same hermitean operator.
But if on top of that you require a measurement to be *ideal* so that the corresponding state that goes with the pointer state |psi1> is the projector P1 applied to the state of the system, THEN the unitary operator that goes with it IS uniquely determined from the Hermitean operator.

cheers,
Patrick.