Quantum State Representation with Commuting Operators

Question

Let $[A,B]=0$. Then, we can find a set of eigenvectors $\{|a_n,b_n\rangle\}$ common to both $A$ and $B$. According to this, and my own understanding, it makes sense to write an arbitrary quantum state as $$\tag{1}|\Psi\rangle=\sum_n \sum_i c_n^i |a_n,b_n,i\rangle,$$ where the sum over $n$ goes over all the eigenvectors, and the sum over $i$ allows for degeneracy to exist.

To me, it seems like we're saying $|a_n,b_n\rangle$ is a single eigenvector common to $A$ and $B$, that could have very well be written as $|w_n\rangle$. This also makes sense.

Yet, Cohen's quantum mechanics text writes $$\tag{2}|\Psi\rangle=\sum_n \sum_p \sum_i c_{n,p,i}\ |a_n,b_p,i\rangle.$$ This has greatly confused me as it seems like we are dealing with two different sets of eigenvectors, one for $A$ and one for $B$. This representation (at least to me) says for each $n$, we are going over all $p$ eigenvectors and account for their degeneracy. Whereas the representation in Eq. (1) says to simply go over the eigenvectors $|a_n,b_n\rangle$ and account for their degeneracy.

Any help in trying to understand where I'm going wrong is appreciated.

Answer 1

If you think of $a_n$ and $b_j$ as eigenvalues, it is quite possible for the eigenvalue $a_n$ to occur more than once, but there is no reason for all eigenstates of $\hat A$ with eigenvalue $a_n$ to have the same eigenvalue of $\hat B$.

An easy example would be eigenstates of the hydrogen atom. The eigenstates $\vert \psi_{n,\ell,m}\rangle$ all have the same energy $E_n$ for fixed $n$, but there are (usually) several states with energy $E_n$ having different eigenvalues of $\hat L^2$ and $\hat L_z$. Hence, an expansion in terms of these eigenstates would contain a different coefficient for each $(n,\ell,m)$ triple of quantum numbers, i.e. $$ \Psi=\sum_{n\ell m}c_{n\ell m}\vert\psi_{n\ell m}\rangle $$

One can even imagine situation where some triples occur more than once, in which case you’d need the extra label $i$ to distinguishing between those states.

Answer 2

In the first equation, n is the index of the pairs ${a_n,b_n}$ of simultaneous eigenvalues, and $i\geqslant 1$. In the Cohen's book equation, he has two indexes, n for $A$ eigenvalues and p for $B$ eigenvalues. If we have a system with four states $|a>$, $|b>$, $|c>$, $|d>$, that have the property that: $A|a>=|a>$, $A|b>=|b>$, $A|c>=3|c>$, $A|d>=3|d>$

$B|a>=0$, $B|b>=2|b>$, $B|c>=4|c>$, $B|d>=4|d>$

In first notation you mention, we will have $n=1,2,3$, and the states will be called $|{a_1,b_1,1}>=|a>$, $|{a_2,b_2,1}>=|b>$, $|{a_3,b_3,1}>=|c>$, $|{a_3,b_3,2}>=|d>$, with $a_1=1, a_2=1, a_3=3, b_1=0, b_2=2, b_3=4$. In the second notation, we will have $n=1,2$, $p=1,2,3$, and the states will be called $|{a_1,b_1,1}>=|a>$, $|{a_1,b_2,1}>=|b>$, $|{a_2,b_3,1}>=|c>$, $|{a_2,b_3,1}>=|d>$.

I think that that's the idea...

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Answer 3

If $A$ and $B$ are commuting self adjoint operators (more precisely operators with pure point spectrum whose spectral measures commute), then the Hilbert space is decomposed into a direct orthogonal sum of common eigenspaces, where $A$ and $B$ are trivially represented as multiplicative operators: $aI$ and $bI$.

The crucial observation is that common eigenspaces of $A$ and $B$ are one-to-one with the pairs of eigenvalues $(a,b)$. Now there are several ways to bijectively label all these couples, i.e., all these common eigenspaces.

In the first case $n$ labels different common eigenspaces whose dimension is the range of possible values of $i$ (which depends on $n$). So it may happen that $a_n= a_{m}$ for example even if $n\neq m$. But $(a_m,b_m) \neq (a_n,b_n)$ necessarily if $n\neq m$ (that is $b_n\neq b_m$ in the considered example).

In the second case $n$ and $p$ separately label different eigenvalues of $A$ and $B$ respectively, and $i$ (whose range depends on $n$ and $p$) accounts for the dimension of every common eigenspace as before. So, in particular, $a_n \neq a_m$ necessarily if $n\neq m$ and $b_p \neq b_q$ if $p\neq q$.

These two procedures to describe the decomposition of the Hilbert space into a direct orthogonal sum of common eigenspaces of $A$ and $B$ are completely equivalent. Using one or the other is just matter of convenience.

A sort of confusion is generated from the absence of indication of the ranges of indices in the sum, especially that of $i$. A less sloppy notation would help understand the equivalence of decompositions.

Answer 4

Using $|a_n,b_p,i\rangle$ is just a more general way to decompose the general state $|\psi\rangle$. Since, all you need is to have a set of states that diagonalize your two operators. For example, if $$A|a_n,b_p,i\rangle=a_n|a_n,b_p,i\rangle \qquad \forall n,p,i$$ and $$B|a_n,b_p,i\rangle = b_p|a_n,b_p,i\rangle\qquad\forall n,p,i$$ then $$[A,B]|a_n,b_p,i\rangle = (a_nb_p-b_pa_n)|a_n,b_p,i\rangle = 0$$

If this set of state is complete (spans the whole Hilbert space) thus you have found what the theorem guarantees exists.

The difference between Eq (1) and (2) is that equation (1) only allows operators spanning the same Hilbert space (because for each eigenvalue of $A$, there is a corresponding eigenvalue for $B$ for the same state). Thus, it would be possible to use only one index for each eigenstate of both operators A and B (like $|a_nb_n\rangle\equiv|n\rangle$ where $A|n\rangle = a_n|n\rangle$ and $B|n\rangle = b_n|n\rangle$). On the contrary, equation (2) allows two operators acting on different Hilbert space (for a single eigenvalue of $A$, there is a whole set of states that gives the same eigenvalue for $A$ but not the same for $B$). This form would be more general.

For example, for a system with a particle that has spin degeneracy (like an electron in an H atom), you could have an operator acting on a Hilbert space spanned by an infinite set of eigenstates (like the Energy operator) that commutes with the spin operator which spans a restricted Hilbert subspace for each energy state (two-fold spin degeneracy for each energy eigenvalue). Thus, your general electron state would be better written by eq (2) since both operators do not act on the same Hilbert subspace. I hope this helps!