In decoherence theory, we try to remove the measurement postulate from Q.M to replace it by unitary evolutions.

Consider a two level system $S$, an apparatus $A$ and the environment $E$.

A first model of measurement is the following :

If $S$ is in $|\psi_S \rangle = \alpha |0\rangle + \beta |1\rangle$, I will have an interaction with the apparatus such that :

$$|\psi_S \rangle |A_i \rangle \rightarrow \alpha |0\rangle |A_0 \rangle + \beta |1 \rangle |A_1\rangle $$

However, this model is not good enough to describe a measurement because, if $\alpha=\beta=\frac{1}{\sqrt{2}}$, I have :

$$\frac{1}{\sqrt{2}} \left( |0\rangle |A_0 \rangle + |1 \rangle |A_1\rangle \right) = \frac{1}{\sqrt{2}}\left( |-\rangle |A_- \rangle + |+ \rangle |A_+\rangle \right) $$

Where I defined $|\pm\rangle = \frac{1}{\sqrt{2}} \left(|0\rangle \pm |1\rangle \right) $ (same idea for the apparatus).

$|0\rangle$ and $|1\rangle$ are eigenstates of $S_z$ spin, and $|+\rangle$ and $|-\rangle$ are eigenstates of $S_x$ spin.

The exact reason why this model is not good are still a little obscure for me (see What was exactly the preferred basis problem in decoherence ).

But what I understood from it is that it is not a good model because with this way of rewriting the vector, one could say "we measured $S_z$", but another one could say "we measured $S_x$".

Then, to solve the problem, we say that actually the apparatus+system also entangle with the environment. And this interaction remove the ambiguity of the measurement basis.

In practice it means that we also have :

$$\frac{1}{\sqrt{2}} \left( |0\rangle |A_0 \rangle + |1 \rangle |A_1\rangle \right)|E_i\rangle \rightarrow \frac{1}{\sqrt{2}} \left( |0\rangle |A_0 \rangle |E_0\rangle + |1 \rangle |A_1\rangle |E_1\rangle \right)$$

And then there is apparently no more the ambiguity basis when we take the partial density matrix of $S+A$, it is written in Chap IV, part A of

Decoherence, einselection, and the quantum origins of the classical

From Zurek.

We have :

$$\rho_{S,A}=\frac{1}{2} \left( |0 A_0\rangle \langle 0 A_0 | + |1 A_1\rangle \langle 1 A_1 | \right) $$

But for me there is still a basis ambiguity.

Indeed, the matrix is the identity in $span(|0 A_0\rangle, |1 A_1 \rangle)$. Then if I write for example : $|\phi_{\pm}\rangle = \frac{1}{\sqrt{2}} \left(|0 A_0 \rangle \pm |1 A_1 \rangle \right) $ for example, I also have :

$$\rho_{S,A}=\frac{1}{2} \left( |\phi_+\rangle \langle \phi_+ | + |\phi_-\rangle \langle \phi_- | \right) $$

Thus, we have a lot of basis in which $\rho_{S,A}$ is diagonal : there is somehow still a basis ambiguity.

However, maybe we are only looking for basis in which the states of the mixture are separable ? And for this case there is only the basis : $(|0A_0\rangle, |1,A_1\rangle)$.

But I am not sure of this.

In summary : how does the interaction with the environment solve the basis ambiguity problem ?


1 Answer 1


Warm-up: pure-state version

The basis ambiguity problem is an artifact of considering only two subsystems, represented by the two factors in $\sum_s c_s |s\rangle|A_s\rangle$. In reality, any useful measurement will compel the thing being measured to influence its surroundings in a prolific way, resulting in a practically irreversible cascade of effects into many uncontrolled parts of the system (the electronics, the table, the atmosphere, and so on). So a more realistic (but still idealized) model of a measurement would be something like $$ \sum_s c_s|s\rangle|A\rangle|B\rangle|C\rangle\cdots \to \sum_s c_s|s\rangle|A_s\rangle|B_s\rangle|C_s\rangle\cdots. \tag{1} $$ In this model, each factor represents some chunk of the system that is complex enough so that different values of $s$ lead to almost-exactly-mutually-orthogonal states of that chunk.

To see why this eliminates the basis ambiguity problem, consider the state $$ |\psi\rangle = |0\rangle|A_0\rangle|B_0\rangle + |1\rangle|A_1\rangle|B_1\rangle, \tag{2} $$ where both terms are understood to be non-zero, and each factor in the first term is understood to be orthogonal to the corresponding factor in the second term. I'll call this a "superposition of factor-by-factor orthogonal terms." The claim is that there is no basis ambiguity in this case. To see why the $0/1$ basis has a unique status, consider some other orthonormal basis $2/3$ and write $$ |0\rangle=a|2\rangle+b^*|3\rangle \hskip2cm |1\rangle=b|2\rangle-a^*|3\rangle. \tag{3} $$ (The coefficients are arranged to respect the orthogonality of $|0\rangle$ and $|1\rangle$.) Let $P$ be the operator that projects the first factor onto $|2\rangle$. Applying $P$ to (2) gives $$ P|\psi\rangle = |2\rangle\otimes \big(a|A_0\rangle|B_0\rangle + b|A_1\rangle|B_1\rangle\big). \tag{4} $$ The factor in parentheses is separable (non-entangled) if and only if either $a$ or $b$ is zero, in which case the $2/3$ basis is just a relabeling of the $0/1$ basis. This shows that the $0/1$ basis has a unique status in the state (2), so there is no basis ambiguity with respect to this factorization of the Hilbert space.

Resolving the basis ambiguity: mixed-state version

The OP asks specifically about the mixed state $$ \rho\propto |0 A_0\rangle\langle 0 A_0| + |1 A_1\rangle\langle 1 A_1|, \tag{5} $$ which is obtained from (2) by tracing over $B$ (which was denoted $E$ in the OP). The $0/1$ basis still has a unique status in the bipartite mixed state (5), just as it did in the tripartite pure state (2), and the proof is similar. To see that the $0/1$ basis still has a unique status, define $P$ as before, and consider the (unnormalized) density matrix $P\rho P$. The result is $$ P\rho P\propto |2\rangle\langle 2|\otimes\Big( |a|^2|A_0\rangle\langle A_0| + |b|^2|A_1\rangle\langle A_1| \Big). \tag{9} $$ This shows that if $a$ and $b$ are both non-zero, then $P\rho P$ is a mixed state; but if either $a$ or $b$ is zero (in which case the $2/3$ basis is just a relabeling of the $0/1$ basis), then $P\rho P$ is a pure state. This shows that the $0/1$ basis has a unique status in the state (5), so there is no basis ambiguity.


As mentioned earlier, we really ought to have lots of tensor factors, not just three and certainly not just two. In the many-partite case, there are no simple (i.e., few-factor) observables that are sensitive to interference between those two terms. In other words, we have no practical way of determining the relative phase of those two terms. In such a situation, we might as well replace the superposition of the factor-by-factor orthogonal terms with an appropriately-weighted mixture of those same terms. This doesn't involve taking any partial trace; it only involves discarding the relative phase information that we aren't able to access with any manageably simple observables anyway. I assume this is where the name "decoherence" came from.

The formalism in which different subsystems are modeled by different tensor factors in the Hilbert space is often convenient, but it has a limitation: it forces us to choose boundaries between those "subsystems." Such choices are clearly artificial. This is not a limitation of quantum theory; it's only a limitation of the (convenient) formalism that tries to use a factorization of the Hilbert space to represent a list of "subsystems."

More importantly, no matter what formalism we use, decoherence theory doesn't solve the measurement problem. It only allows us to recognize when (and to what degree) something has been "measured," so that we know when we can safely use Born's rule.

  • $\begingroup$ Here you showed that for 3 systems, we have a unique basis to define $|\psi\rangle$ as a "factor by factor [...]", and thus this remove the basis ambiguity. I agree that it removes the basis ambiguity for all of the 3 systems. But then I will explain my exact question more accurately. I thought that the basis ambiguity was about when I focus on system+apparatus. That we have different ways to interpret their state. Consider the two ways I wrote $\rho_{S,A}$ in my post. Isn't it a problem that we have different way to write this density matrix ? $\endgroup$
    – StarBucK
    Commented Mar 10, 2019 at 20:15
  • $\begingroup$ So maybe my question is actually all remaining in "why do we have a basis ambiguity" ? I agree with the math that for two system I will not have a unique decomposition in the opposite of for three system as you showed the uniqueness. But then, what are the physical conclusion of this remark ? Why would this uniqueness "save" us ? What was the physical problem we had before that we don't have after ? $\endgroup$
    – StarBucK
    Commented Mar 10, 2019 at 20:23
  • 1
    $\begingroup$ @StarBucK I deleted my earlier comments and edited the answer to focus on the question about $\rho_{S,A}$. The answer still starts with the pure-state version, but with a simplified proof whose notation is re-used to prove the corresponding result in the mixed state case. $\endgroup$ Commented Mar 10, 2019 at 22:28
  • 1
    $\begingroup$ @StarBucK Yes, that's right. Decoherence theory doesn't quite say that the state will collapse, but it says that we might as well go ahead and discard all but one of the terms (whichever one we experience), because we'll never be able to detect any interference with the other terms even if they still exist. So, like you said, the measurement problem is still a problem, but quantum theory can at least tell us which observable is measured -- that is, into which basis the state might as well have "collapsed" (as far as we can ever tell). $\endgroup$ Commented Mar 11, 2019 at 0:54
  • 1
    $\begingroup$ @DvijD.C. Here's an example of the distinction I had in mind: consider a tripartite system in the pure state $$ a|000\rangle+b|111\rangle,$$ where $0$ and $1$ indicate orthogonal states of each subsystem. If the theory doesn't have any practically-measurable observables that mix the two terms in this superposition, then we might as well represent it by the density matrix $$|a|^2|000\rangle\langle 000|+|b|^2|111\rangle\langle 111|$$ instead. This doesn't involve taking a partial trace, it simply discards the practically-inaccessible relative phase information. $\endgroup$ Commented Aug 4, 2020 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.