# Non Commuting Observables not Representing a Complete Set?

1. May 31, 2014

### kq6up

I am going through James Binney's course on Quantum Mechanics. I love all of the little misconceptions he points out along the way. One thing he mentions in his text and the lectures is found on page 20 and 21 starting with the heading "Commutators" eq. 2.21. He states that non commuting observables do not form a complete set. Could someone unpack this idea for me, or point me to some text that expands on this idea a little more clearly?

Here is a link to his text book. It is FREE:

http://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf

Thanks,
Chris Maness

2. May 31, 2014

### micromass

It says that there is no complete system of mutual eigenkets of $A$ and $B$.

So since $A$ and $B$ are self-adjoint, they have both a complete system of eigenkets. In LA language, it is said that we can diagonalize $A$ and $B$. But we cannot do it simulatenously. So if we have a complete system of eigenkets of $A$, then some of them won't be eigenkets to $B$.

3. May 31, 2014

### hilbert2

It means that if the operators $A$ and $B$ do not commute, it is impossible to find a set of states $\left|\phi_i\right>$ such that:

1. All states $\left|\phi_i\right>$ are eigenstates of both $A$ and $B$.
2. Any imaginable state $\left|\psi\right>$ can be written as a linear combination of the states $\left|\phi_i\right>$. This is what it means if a set of vectors forms a complete set in the vector space being considered.

EDIT: Micromass was faster than me...

4. May 31, 2014

### Jilang

He says that if A and B don't commute it is impossible to find a complete set of mutual eigenkets of A and B. This means that it is impossible to have an eigenket of A and an eigenket of B. As a consequence this means that you cannot have a state that has a definite value of A and a definite value of B.

Edit. Everyone is faster than me!

5. May 31, 2014

### hilbert2

^ Actually, that is not true. As pointed out in the text linked by the OP, it is possible that some ket is "accidentally" an eigenket of two mutually non-commuting observables. It's just that one can never find enough such kets to span the whole vector space.

6. May 31, 2014

### kq6up

Ahhh, I like that. Self adjoint meaning -- hermitian. That is, when the eigenvectors are orthogonal -- which in QM they always are (I think).

Chris

7. May 31, 2014

### Matterwave

Self-adjoint and Hermitian are very similar, but not exactly the same. Self-adjoint is a slightly stronger condition that requires the domain of the two operators (the operator itself and its adjoint) to be identical.

Eigenvectors of distinct eigenvalues of self-adjoint operators are orthogonal. Degenerate eigenvectors need not be orthogonal, but one can use the Graham-Schmidt procedure to make them so.

8. May 31, 2014

### dextercioby

Think about the harmonic oscillator in 1D. The Hamiltonian doesn't commute with neither momentum, nor coordinate. The eigenvectors of the Hamiltonian don't even live in the same space as the eigenvectors of either momentum or coordinate. Now go to the occupation number formalism. The number operator's eigenvectors are eigenvectors for the Hamiltonian (since H = N + 1/2). H and N will then commute.

9. May 31, 2014

### kq6up

Ok, I have had a little time to scratch my head on this one. In general operators (matrices) do not share eigenkets (eigenvectors). So is there some positive relationship between having common eigenket and commutation? In other words -- if two operators share eigenkets, they commute, and if not they don't?

This seems like a stretch, since I am at a loss as to how to see if this is true with some sort of proof.

Thanks,
Chris Maness

10. May 31, 2014

### kq6up

Ok, I think I can actually show what I just said. Let-a-rip:

$${ C }^{ -1 }QC=D\quad and\quad let\quad Q=AB$$ Where A & B are hermitian. That is:

$$A={ A }^{ \dagger }\quad and\quad B={ B }^{ \dagger }$$

Let C be the unitary matrix that diagonalizes Q.

$${ C }^{ \dagger }ABC=D\quad implies\quad { \left( { C }^{ \dagger }ABC \right) }^{ \dagger }={ D }^{ \dagger }=D\quad \therefore \quad D={ C }^{ \dagger }BAC$$

This seems to imply AB=BA under the conditions above. I know it doesn't show that C the unitary matrix that diagonalizes (spectrally decomposes) A and B separately. I think I would have to show that to be sure.

Is this any good?

Thanks,
Chris Maness

11. May 31, 2014

### micromass

For one direction (for matrices at least), see http://www.math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect2.pdf Section $3$. So this shows that any family of commuting normal operators are simultaneously diagonalizable.

As for the other direction. Let $A$ and $B$ be simulatenously diagonalizable. Then there is (by definition) a unitary matrix $C$ and diagonal matrices $D$ and $D^\prime$ such that $CAC^{-1} = D$ and $CBC^{-1} = D^\prime$. But then $CABC^{-1} = DD^\prime = D^\prime D = CBAC^{-1}$. It follows that $AB = BA$.

As for what you did, you assumed that there is a real diagonal matrix $D$ such that $CABC^{-1} = D$. You cannot just assume this. This assumption is actually equivalent to asking that $AB$ is self-adjoint and that is only the case if $AB=BA$ (it is even equivalent to this).

12. May 31, 2014

### kq6up

Thanks, micromass. It is now clear to me why what you shared with me is true. I think you meant to put the ^(-1) on the right "C". I would still like to get better physical insight into what he stated. I get that an abstract space can be covered (I take that to mean that you can add the basis together in such a way to reach every point in Hilbert/Abstract space can be reached by this set of basis). However, why this would be the case for eigenkets of operators that don't commute might still be eluding me. Is it because there are not enough kets/vectors to span all space since a non commuting pair would not necessarily produce a full basis set of kets to span this space?

Thanks,
Chris Maness

13. Jun 1, 2014

### kq6up

Anyone still around to verify my last statement?

Thanks,
Chris Maness

14. Jun 1, 2014

### Matterwave

A given Hermitian operator will have a complete basis of eigen-vectors. Two Hermitian operators will have a complete basis of mutual eigen-vectors, this is because if $A\left|a_n\right>=a_n\left|a_n\right>$ then the $\left|a_n\right>$ are complete, and given $AB=BA$ then:

$$A(B\left|a_n\right>)=B(A\left|a_n\right>)=a_n(B\left|a_n\right>)$$

This means that although B transformed the state $\left|a_n\right>$ into another state $B\left|a_n\right>$ this transformed state is still the eigen-vector of $A$ with eigenvalue $a_n$. Given the case of non-degenerate eigen-vectors where each eigenvalue corresponds to only one eigen-vector (for a more complete proof, where degeneracies are considered, one can look at e.g. Ballentine) then this means that $B\left|a_n\right>=b_n\left|a_n\right>$ for some constant $b_n$. So the eigen-vectors of $A$ are also eigen-vectors of $B$.

The fact of completeness is much harder to prove, and is given basically by the spectral theorem. If A and B do not commute, then a given eigen-vector of A will not necessarily be an eigen-vector of B. The eigen-vectors of A and B both still form a complete set, but they no longer coincide with each other.