# Uncertainity question

1. Dec 29, 2009

### chafelix

Interesting problem: Suppose one has 2 hermitian operators, O1,O2 with distinct eigenvalues.
Say the first is the Hamiltonian. We measure the energy, get a value, say E_1. So the system
is in |1>. Suppose now that the second operator has the property to turn state 1 into state |2> and operating on |2> gives |1>
Hence if after having measured the energy to be E_1, we operate on the state with the second op, we get |2>. No uncertainty about the outcome.
So should not the uncertainity be 0?
So O2|1>=|2>,<1|O2|1>=0
O2**2|1>=O2|2>=|1>. So <1|O2**2|1>=<1|1>=1
But uncertainity is (<1|O2**2|1>-<1|O2|1>**2)=1, not 0.

2. Dec 29, 2009

### sokrates

Could you elaborate on how you define 'uncertainty' ?

I thought it was about two different non-commuting observables, but you describe the energy here only?

What is the other operator here?

3. Dec 29, 2009

### Fredrik

Staff Emeritus
Why do you think the uncertainty of O2 in state |1> should be 0? |1> is not an eigenstate of O2.

4. Dec 30, 2009

### chafelix

Uncertainity is sqrt(expectation value of operator**2)-square of expectation value)
So: Expectation value of operator O2 is <1|O2|1>=0
Expectation value of O2**2 is 1

5. Dec 30, 2009

### Fredrik

Staff Emeritus
Yes, and that means that the uncertainty of O2 in state |1> is 1. Why do you think it should be 0?

6. Dec 31, 2009

### chafelix

If any experiment can only give a single result, then there is no uncertainity.
In this case where we start with a prepared state |1>, a measurement of O2 can only give the result |2>. It has 100% |2> character and 0% |1> character.
In contrast if O2|1> were a|1>+b|2>, then the outcome would be uncertain, with respective probabilities to give |1> or |2> that are |a|**2 and |b|**2

The answer is that |1> and |2> are not eigenvectors of O2, so what one should do is use a different basis that is the basis of the O2 eigenvalues, Remember, the possible outcomes are only eigenvalues of the operator measured(here O2) and the state the system is left in is an eigensate of O2. If you do that , then the new basis is of the form
|P>=c|1>+d|2>, so that O2|P>=p|P>=>c|2>+d|1>=pd|2>+pc|1>=>
pd=c,pc=d=>p=+-1, so that a state |2> is not an eigenstate of O2, but a linear combination of its eigenstates (|1>+|2>)/sqrt(2) and (|1>-|2>)/sqrt(2)

7. Dec 31, 2009

### Fredrik

Staff Emeritus
That's right. (The second half of your post. I assume the first half is just what you were thinking before). The uncertainty relation for these operators and the state |1> is 0·1≥0. Well, I haven't actually calculated the right-hand side, but it would have to be 0, since the left-hand side is 0.