Proof of Commutative Operators for Simultaneous Measurement

[VVS2000]

TL;DR

Had a confusion about commutative operators

I was seeing a lecture and the professor told that
i) you can measure an operator if it's hermitian, therefore observables are hermitian operators
ii) if you can measure two observables simultaneously, then those two observables(operators)
Is there any proof for this or is it some kind of rule?
Thanks in advance..

 

[PeroK]

Summary:: Had a confusion about commutative operators

I was seeing a lecture and the professor told that
i) you can measure an operator if it's hermitian, therefore observables are hermitian operators
ii) if you can measure two observables simultaneously, then those two observables(operators)
Is there any proof for this or is it some kind of rule?
Thanks in advance..

You mean a proof that:

i) Observables are represented by Hermitian operators?

ii) Compatible observables are represented by commuting (Hermitian) operators?

 

[VVS2000]

You mean a proof that:

i) Observables are represented by Hermitian operators?

ii) Compatible observables are represented by commuting (Hermitian) operators?

Yes

 

[PeroK]

Yes

(i) is usually taken as a postulate of QM.

(ii) follows from the fact that if two observables are compatible, then their operators need a shared eigenbasis. It's not too hard to prove that for commuting operators, you can find a shared eigenbasis.

How much linear algebra do you know?

 

[VVS2000]

(i) is usually taken as a postulate of QM.

(ii) follows from the fact that if two observables are compatible, then their operators need a shared eigenbasis. It's not too hard to prove that for commuting operators, you can find a shared eigenbasis.

How much linear algebra do you know?

Well I know some basics about vector spaces, I know what a hermitian operator is and what are bras and kets and inner product and all that. I am not able to make sense out of why they have to be hermitian and why does is commutativity indicate that you can measure them simultanueously

 

[PeroK]

Well I know some basics about vector spaces, I know what a hermitian operator is and what are bras and kets and inner product and all that. I am not able to make sense out of why they have to be hermitian and why does is commutativity indicate that you can measure them simultanueously

The operators are taken to be Hermitian as a postulate. The motivation is that Hermitian operators a) have real eigenvalues (and the eigenvalues are the possible results of measurements) and b) their eigenvectors form a basis.

Two observables being compatible means that if you measure one, then the other this does not disturb the first measurement. This means that they must have a shared eigenbasis, as a measurement throws the state into an eigentstate of the observable.

Measure observable $A$; system is left in an eigenstate of $A$; measure $B$; if $B$ is compatible with $A$, then the state does not change at this point, so it must also be an eigenstate of $B$. So, compatible observables must have common eigenstates.

The proof that commuting operators have common eigenstates is easily shown if we assume non degeneracy:

Suppose operators $A$ and $B$ commute and let $Av = \lambda v$, where $v$ is an eigenvector of $A$ with eigenvalue $\lambda$. Now

$A(Bv) = B(Av) = B(\lambda v) = \lambda(Bv)$

So, $Bv$ is also an eigenvector of $A$ with eigenvalue $\lambda$. Hence, in the case of non-degeneracy $Bv$ is a scalar multiple of $v$ and $v$ is also an eigenvector of $B$. QED

The case of degenerate eigenvalues is a little trickier.

 

[PeterDonis]

The proof that commuting operators have common eigenstates

But what you actually need to prove is the converse: that if the operators have common eigenstates, they must commute.

 

[PeroK]

But what you actually need to prove is the converse: that if the operators have common eigenstates, they must commute.

You have to prove it both ways. With appropriate assumptions that's the easy case, as any vector can be expressed in the shared eigenbasis.

The meat of the proof is the converse.

 

[VVS2000]

The operators are taken to be Hermitian as a postulate. The motivation is that Hermitian operators a) have real eigenvalues (and the eigenvalues are the possible results of measurements) and b) their eigenvectors form a basis.

Two observables being compatible means that if you measure one, then the other this does not disturb the first measurement. This means that they must have a shared eigenbasis, as a measurement throws the state into an eigentstate of the observable.

Measure observable $A$; system is left in an eigenstate of $A$; measure $B$; if $B$ is compatible with $A$, then the state does not change at this point, so it must also be an eigenstate of $B$. So, compatible observables must have common eigenstates.

The proof that commuting operators have common eigenstates is easily shown if we assume non degeneracy:

Suppose operators $A$ and $B$ commute and let $Av = \lambda v$, where $v$ is an eigenvector of $A$ with eigenvalue $\lambda$. Now

$A(Bv) = B(Av) = B(\lambda v) = \lambda(Bv)$

So, $Bv$ is also an eigenvector of $A$ with eigenvalue $\lambda$. Hence, in the case of non-degeneracy $Bv$ is a scalar multiple of $v$ and $v$ is also an eigenvector of $B$. QED

The case of degenerate eigenvalues is a little trickier.

Is that why we can only measure one component of a spin at a time? Because spin operators are not commutative?

 

[PeroK]

Is that why we can only measure one component of a spin at a time? Because spin operators are not commutative?

Essentially, yes.

 

[VVS2000]

Thanks a lot

 

[hilbert2]

A pair of non-commuting hermitian operators can have some common eigenstates, but not a whole basis of them. For instance, any hydrogenic s-orbital is an eigenstate of $L_x$, $L_y$ and $L_z$ with eigenvalue $0$.

 

[vanhees71]

The statement that "you can measure a hermitian operator" is strange, because to measure some thing I first have to get hold of it in the real world. I've never touched an operator in a Hilbert space. It's a concept i the mind of mathematicians and theoretical physicists used to formulate QT, but not something you can find as a real thing in Nature.

 

[Isaac0427]

Here's a slightly different perspective on this.

i)
If we want to have operators representing observables, which is a postulate of quantum mechanics, those operators would have to be Hermitian. This is because of three properties of Hermitian operators that turn out to be necessary for representing an observable:
a) Eigenvalues of Hermitian operators are real. Since the eigenvalues represent the physical value of the observable, it is obvious that these have to be real.
b) Eigenstates of Hermitian operators with different eigenvalues are orthogonal. Take the example of energy, with two eigenstates with different energies: $\hat{H}\left|\psi_1 \right>=E_1\left|\psi_1 \right>$, $\hat{H}\left|\psi_2 \right>=E_2\left|\psi_2 \right>$, $E_1\neq E_2$. Remember that if a particle is in the state $\left|\Psi \right>$, upon measurement, the probability of getting the energy $E_1$ is $\left|\left<\psi_1 |\Psi\right>\right|^2$. Say a particle is in the state $\left|\psi_2 \right>$. The probability of finding the energy to be $E_1$ must be zero (as the particle definitely has an energy $E_2$), and thus $\left|\left<\psi_1 |\psi_2\right>\right|^2=0$ or $\left<\psi_1 |\psi_2\right>=0$.
c) Any state can be represented purely as a superposition of eigenstates of a Hermitian operator (i.e. eigenstates of Hermitian operators form a complete basis). The physical statement of this for energy is that any particle's state is some combination of states with allowed energies. If we couldn't express a particle's state this way, then there would a possibility for the particle not to have one of the allowed energies.

There's an even easier explanation of this, though. We know the expectation value of an observable Q with an operator $\hat{Q}$ is $\left<Q\right>=\left<\psi |\hat{Q} \psi \right>$. Since the expectation value must be real, $\left<Q\right>=\left<\psi |\hat{Q} \psi \right>$ must be the same as $\left<Q\right>^*=\left<\hat{Q}\psi |\psi \right>$ meaning $\hat{Q}$ must be Hermitian as $\left<\psi |\hat{Q} \psi \right>=\left<\hat{Q}\psi |\psi \right>$.ii)
This one is shorter. The generalized uncertainty principle says that
$$\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2i}\left<\left[\hat{A},\hat{B}\right]\right>\right)^2.$$
But for compatible observables the LHS is zero, so the RHS must also be zero. Namely $\left<\left[\hat{A},\hat{B}\right]\right>=0$. In fact, in some special cases the observables can be compatible even when the commutator is not zero, as it is just the expectation value of the commutator that must be zero: for example, the state $\left| l,m\right>=\left|0,0\right>$, which is simultaneously an eigenstate of angular momentum in all three directions with an eigenvalue of zero (you can do the same with a spin-0 particle).

I hope this helps.

 

[vanhees71]

Very nice summary. It's only important to keep in mind that everywhere, where you wrote "Hermitian" you should write "self-adjoint", i.e., it's also important that the operators should have a densely defined domain which also is the co-domain of the operator. For a pedagogical review on this issue, see

G. Bonneau and J. Faraut, Self-adjoint extensions of operators
and the teaching of quantum mechanics, Am. Jour. Phys. 69,
322 (2001), https://arxiv.org/abs/quant-ph/0103153.