# Confused about eigenstates, uncertainties,expectation values

1. Mar 20, 2015

### dyn

If a wavefunction can be in different states each with a different eigenvalue when operated on by an Hermitian operator then am I correct in thinking that the expectation value of that operator when acting on a superposition of states could give a value that does not equal any single one of the individual eigenvalues ?
If I am correct so far ; now comes my main question - I have read that the uncertainty is zero if the state is in an eigenstate ,ie for an operator A the uncertainty is zero implies A(psi) = <A>(psi) where <A> is the expectation value of A. What does this mean ?
Thanks.

2. Mar 20, 2015

### Staff: Mentor

To prevent this type of confusion you really need to see an axiomatic treatment.

I will take the two axioms QM is based on from Ballentine with a slight tweak that exposes it a bit better.

It does require a bit of knowledge of linear algebra though - specifically the spectral theorem.

First Axiom:
An observation with outcomes i is described by a resolution of the identity Ei such that the probability of outcome is determined only by the Ei.

We can associate an arbitrary real number yi with each outcome and form a Hermitian operator O = Σ yi Ei to form what called the observations observable. The possible outcomes of the observation are its eigenvalues yi. From the spectral theorem given O we can uniquely recover the Ei and yi so either view is equivalent.

Second Axiom - Called The Born Rule
There exists a positive operator P of unit trace, called the state of the system, such that the expected outcome of the observation O is Trace (PO).

The second axiom is not entirely independent of the first because of a deep and important theorem called Gleason's theorem - but that is just by the by.

Note a state is a positive operator. But positive operators of the form |u><u| are, without going into the detail, called pure and they are what you usually deal with starting out in QM.

The Born Rule for pure states is E(O) = <u|O|u>.

Now lets suppose the state is in an eigenstate of an observable O = ∑ yi |bi><bi|. Since the yi are arbitrary we can change it to all zero except for yi =1 ie O = |bi><bi|. This would mean the observation gives 1 if outcome i occurs and zero otherwise. Then E(|bi><bi) = probability of outcome i occurring. Thus if the system is in state |bi> the probability of outcome i is <bi|bi><bi|bi> = 1. Thus if the system is in a state corresponding to an eigenvalue of the observable, the outcome of the observation will be a dead cert - it will always give the yi associated with that state in the observable.

Unfortunately the explanation does require knowledge of the Bra-Ket notation and linear algebra. I wish I could explain it without it - but I really can't.

Now specifically to your question. The outcome of the observation must be one of the eigenvalues of the observable - it cant be anything else. But of course the expected value can be a lot different - its similar to saying the average number of children in a household is 2.5 - you can't of course have .5 of a child.

Thanks
Bill

Last edited: Mar 20, 2015
3. Mar 20, 2015

### dyn

Thanks for that but its a bit over my head. I was thinking of a simple example like momentum eigenstates +p and -p but the expectation value is zero which cannot be the result of a single measurement.

4. Mar 20, 2015

### Staff: Mentor

You might not have seen the end bit of my post - I sometimes do it in parts.

From the very axioms of QM only the eigenvalues of the observable are possible outcomes of an observation. But when you take the average you can get numbers a lot different to those outcomes - like the .5 of a child.

Thanks
Bill

5. Mar 21, 2015

### dyn

thanks for that. So the first part of my OP was correct so what ( if possible in easy terms) are the eigenstates that lead to zero uncertainty ? Is it each individual one ? But that wouldn't explain the expectation value appearing in the equation I quoted ?

6. Mar 21, 2015

### Staff: Mentor

If you observe a system that is in an eigenstate of the observable you will get the eigenvalue of the eigenstate 100% for sure. There is no uncertainty in the outcome. That, as I tried to explain in the bit that was a bit over your head (sorry - but I don't know any other way to explain it), is a consequence of the very axioms of QM.

It is in fact a simple consequence of the expectation formula E(O) = <u|O|u> - again as I tried to explain.

Thanks
Bill

7. Mar 21, 2015

### Fredrik

Staff Emeritus
Yes. However, it's the physical system that can be in different states, not the wavefunction. The wavefunction is the state.

To say that f is an eigenvector of A is to say that there's a number $a$ such that $Af=af$. So if f is an eigenvector of A, there's a number $a$ such that the expectation value of A in state f is
$$\langle A\rangle_f =\langle f,Af\rangle=\langle f,af\rangle=a\langle f,f\rangle =a.$$ Then the uncertainty of A in state f is
$$(\Delta A)_f =\sqrt{ \Big\langle \big(A-\langle A\rangle_f\big)^2 \Big\rangle_f } =\sqrt{ \Big\langle \big(A-a^2\big)^2 \Big\rangle_f } =\sqrt{ \langle A^2-2aA+a^2 \rangle_f } =\sqrt{a^2-2a^2+a^2} =0.$$ Note that a notation like T+t where T is an operator and t is a number really means $T+tI$, where $I$ is the identity operator.

You asked specifically about the equality A(psi) = <A>(psi). In my notation, we have
$$Af=af=\langle A\rangle_f f.$$ This can't be simplified to $A=\langle A\rangle_f$, since $\langle A\rangle_f$ depends on f. (If B is an operator and b is a number such that $Bf=bf$ for all f, then we can conclude that $B=bI$, where $I$ is the identity operator, and we can use the convention to write the right-hand side simply as $b$).

Last edited: Mar 21, 2015
8. Mar 21, 2015

### dyn

Thanks. I think I get that but what happens when the state is a superposition of kets ? Can the state still be an eigenvector ? and what would be the expectation value be that would lead to zero uncertainty ?

9. Mar 21, 2015

### Fredrik

Staff Emeritus
A superposition of kets (=state vectors = wavefunctions) is still a ket. Maybe you meant to ask about a superposition of eigenkets? A linear combination of any number of eigenvectors all with the same eigenvalue λ is an eigenvector with eigenvalue λ. But a linear combination (with non-zero coefficients) of eigenvectors with different eigenvalues isn't an eigenvector.

If you mean that there should be a number s such that if $\langle A\rangle_f=s$, then $(\Delta A)_f=0$, I don't think there is such a number, except maybe when we're dealing with some very special A.

The uncertainty is zero if and only if its square is zero. If $(\Delta A)_f^2=0$, we have
\begin{align}
&0=\langle A\rangle_f^2 =\langle (A-\langle A\rangle_f)^2\rangle_f =\langle A^2-2\langle A\rangle_f A+\langle A\rangle_f^2\rangle_f\\
&=\langle A^2\rangle_f-2\langle A\rangle_f^2+\langle A\rangle_f^2 =\langle A^2\rangle_f -\langle A\rangle_f^2,
\end{align} which implies that
$$\langle A\rangle_f^2=\langle A^2\rangle_f.$$ So we have $(\Delta A)_f=0$ if and only if A and f satisfy the last equality above.

10. Mar 21, 2015

### dyn

Thanks a lot for your help both of you. Quantum mechanics is great. Sometimes I think I get it then I find out I didn't really get it : then I think I get it again !