Is the 2-qubit singlet state invariant under bilateral projections?

[QuantumExplorer]

The 2-qubit singlet state $\ket{\Psi^-} = \frac{1}{\sqrt{2}}(\ket{01}-\ket{10})$ is invariant under bilateral rotations(https://arxiv.org/pdf/quant-ph/9511027), that is any operator of the form $U \otimes U$ for $U$ being unitary. However, I'm wondering if the state is invariant also for a more general transformation $A \otimes A$. In general that transformation will lead to $(A \otimes A )\ket{\Psi^-} = \det(A)\ket{\Psi^-}$. Given this, would it be OK to say that $\ket{\Psi^-}$ is invariant for any $A$, or do we have to worry about the cases when $A$ is not invertible? If $A$ is singular, say $A = \ket{0}\bra{0}$, then in this case $\det(A)=0$ and the application of $A \otimes A$ will result in null vector, which seems different enough from the original to still call the case "invariant".

 

[QuarkyMeson]

$(A \otimes A )\ket{\Psi^-} = \det(A)\ket{\Psi^-}$ in the above context is always true. The proof is trivial.

For invariance you need to define exactly what that means. I briefly looked at the linked paper and I don't fully understand it, but it looks like they're exploiting SU(2) rotations (or maybe also U(2) rotations and ignoring the global phase), which by definition have det = 1. So no det(A) = 0 doesn't make much physical sense in that context and in general a non-unitary A isn't trace preserving on the density matrix.

 

[QuantumExplorer]

$(A \otimes A )\ket{\Psi^-} = \det(A)\ket{\Psi^-}$ in the above context is always true. The proof is trivial.

For invariance you need to define exactly what that means. I briefly looked at the linked paper and I don't fully understand it, but it looks like they're exploiting SU(2) rotations (or maybe also U(2) rotations and ignoring the global phase), which by definition have det = 1. So no det(A) = 0 doesn't make much physical sense in that context and in general a non-unitary A isn't trace preserving on the density matrix.

Do you have any references you'd recommend for types or definitions of invariances? Perhaps you can take a look at the discussion I had about it in the comments in this stack exchange
https://quantumcomputing.stackexcha...rences-between-the-four-different-bell-states.

 

[QuarkyMeson]

I mean, you just kind of make it up and pick a definition for it so everyone is on the same page. You could argue for exact invariances, such that $A\in SU(2)$ or $A\in U(2)$ up to global phase (like the paper). You could say it's invariant up to some scalar (when I hear this I normally assume scalar != 0), or that there is invariance if the span is preserved (this appears to be what Adam Zalcman is arguing).

• Singlet is invariant under A⊗A up to multiplication by a scalar for every A. In fact, we know the scalar: it is detA. This is true for singular A as well, albeit in this case the scalar is detA=0, so while the mathematical invariance holds (after all, the zero vector belongs to the one-dimensional space spanned by the singlet), the resulting vector has no physical meaning as a quantum state (only normalizable vectors do). See equation (9) and surrounding text.
  – Adam Zalcman

This is correct mathematically and for the invariance Adam is arguing for. The resulting vector has no physical meaning as a quantum state, as he said.

Some things I feel he's implying I disagree with, a nonunitary $A \otimes A$
is not a deterministic evolution of a qubit. If $A \otimes A$ is singular and $(A \otimes A )\ket{\Psi^-} = 0$, that does not mean the state becomes the zero vector. It means that particular outcome has probability p=0, it never happens. So I'm fine with what he said to there. However, In quantum optics, using matched polarizers causes a non-trace-preserving operation because any photon whose polarization does not align with the polarizer's transmission axis is absorbed or reflected, effectively resulting in a loss outcome. These outcomes (photon passes vs. photon is lost) are distinguishable by monitoring the coincidence rates of photons. So it's not fair to say physically that $A \otimes A$ is like $U \otimes U$, or that $A \otimes A$ never takes the state out of the singlet subspace.

Physically, I would never define invariance as span preservation for quantum states. I could certainty be mistaken, I'm not an expert here.

 

[QuantumExplorer]

I mean, you just kind of make it up and pick a definition for it so everyone is on the same page. You could argue for exact invariances, such that $A\in SU(2)$ or $A\in U(2)$ up to global phase (like the paper). You could say it's invariant up to some scalar (when I hear this I normally assume scalar != 0), or that there is invariance if the span is preserved (this appears to be what Adam Zalcman is arguing).

This is correct mathematically and for the invariance Adam is arguing for. The resulting vector has no physical meaning as a quantum state, as he said.

Some things I feel he's implying I disagree with, a nonunitary $A \otimes A$
is not a deterministic evolution of a qubit. If $A \otimes A$ is singular and $(A \otimes A )\ket{\Psi^-} = 0$, that does not mean the state becomes the zero vector. It means that particular outcome has probability p=0, it never happens. So I'm fine with what he said to there. However, In quantum optics, using matched polarizers causes a non-trace-preserving operation because any photon whose polarization does not align with the polarizer's transmission axis is absorbed or reflected, effectively resulting in a loss outcome. These outcomes (photon passes vs. photon is lost) are distinguishable by monitoring the coincidence rates of photons. So it's not fair to say physically that $A \otimes A$ is like $U \otimes U$, or that $A \otimes A$ never takes the state out of the singlet subspace.

Physically, I would never define invariance as span preservation for quantum states. I could certainty be mistaken, I'm not an expert here.

Could you say a bit more about the evolution being not deterministic? If I have the singlet state and pass it through matched polarizers, the photons will get absorbed with a 100% chance (equivalently I can say there is a 0% chance they will make it through), but that is a deterministic behavior.

Also, if it is not the zero vector, what does the singlet evolve/ gets collapsed into? Is it the vacuum state, that is a state with zero photons? Is the vacuum state within the span of the singlet? This is a bit puzzling because it seems like the condition of the field after applying $A \otimes A$ will be indistinguishable from me not turning on the source at all. I'm wondering if it is confusing because when I write the singlet as $\ket{\Psi^-} = \frac{1}{\sqrt{2}}(\ket{01}-\ket{10})$ I'm only labeling the polarization direction degree of freedom, but that one becomes effectively undefined after the projection. To make sense of this I need to keep track of both the polarization $\vec{\epsilon}$ and the photon number $n$ as in $\ket{\vec{\epsilon},n}$, so that $\ket{\Psi^-} = \frac{1}{\sqrt{2}}(\ket{H,V,1,1}-\ket{V,H,1,1})$ with each ket being $\ket{\vec{\epsilon}_1,\vec{\epsilon}_2,n_1,n_2}$ and

$$\require{enclose}\enclose{horizontalstrike}{A \otimes A\ket{\Psi^-} = \frac{1}{\sqrt{2}}(\ket{\vec{0},\vec{0},0,0}-\ket{\vec{0},\vec{0},0,0}) \propto \ket{\vec{0},\vec{0},0,0} \neq 0}$$

although I'm still a bit iffy about the above equation, it just seems like you need to subtract $\ket{\vec{0},\vec{0},0,0}$ from itself but you still don't get just $0$.

 

[QuantumExplorer]

After thinking about this a bit more, I realized I made a mistake above. Assuming $A=\ket{H}\bra{H}$:

$$ \begin{align} \nonumber
(A\otimes A )\ket{\Psi^-} &= \frac{1}{\sqrt{2}}\left[(A\otimes A) \ket{H,V,1,1}-(A\otimes A) \ket{V,H,1,1}\right]\\ \nonumber

&= \frac{1}{\sqrt{2}}\left[\ket{H,\vec{0},1,0}-\ket{\vec{0},H,0,1}\right] \\ \nonumber

&\neq 0\ket{\Psi^-} \neq \ket{\vec{0},\vec{0},0,0}
\end{align}
$$

Which solves my issue with subtracting the two branches because they are actually different. So indeed it is not the vacuum, different from me leaving the source off, and it is not deterministic because you will still have one photon circulating through, but it is random on which side it will be. Passing through the matched polarizers will lead to no coincidences because $|\bra{1,1}(A \otimes A) \ket{\Psi^-}|^2 = 0$. But now I'm a bit confused because in this case it seems like $(A \otimes A) \ket{\Psi^-}\neq \det(A)\ket{\Psi^-}$.

 

[QuarkyMeson]

In QM we can look at two different update rules:
$$|\psi\rangle \mapsto U|\psi\rangle$$
with $U$ unitary. This evolution of the state vector is deterministic and preserves norm:
$$\|U|\psi\rangle\|=\||\psi\rangle\|$$
Alternatively, for say a measurement, each possible outcome is associated with an operator $A$ that salsifies $A^{\dagger}A \leq \mathbb{1}$, the outcome occurs with probability:
$$p=\|A|\psi\rangle\|^{2}$$
and if it happens, the (normalized) output state is:
$$|\psi'\rangle=\frac{A|\psi\rangle}{\sqrt{p}} \qquad\text{(only defined if }p>0\text{)} $$

The distinction between these updates is rooted in information conservation. Unitary evolution rotates the state vector without changing its length, ensuring reversibility. In contrast, say a measurement operator $A$ acts as a projection. It effectively deletes the components of the superposition that contradict the observed outcome and because multiple initial states could theoretically map to the same post-measurement state or zero, the mapping is not injective, and therefore not invertible. This loss of information renders the process irreversible and the outcome probabilistic. Getting $|\psi\rangle$ back from $|\psi'\rangle$ is impossible in this case.

As an aside and not to go too far off into lala land but there are also incomplete measurements that are not fully deterministic for the entire system but still preserve a probabilistic reversibility. (This is an invertible non-unitary A, see: https://qubit.guide/4.4-example-of-an-incomplete-measurement)

Could you say a bit more about the evolution being not deterministic? If I have the singlet state and pass it through matched polarizers, the photons will get absorbed with a 100% chance (equivalently I can say there is a 0% chance they will make it through), but that is a deterministic behavior.

So for your example, if I have the singlet state and pass it through matched vertical polarizers, and we look specifically for coincidence counts, the probability is 0.

However, if I tell you no coincidence occurred, can you tell me with certainty that the input state was the singlet state? No. The input could have been the state $|HH\rangle$, which also yields zero coincidences. Even though the outcome appears deterministic, the state vector's evolution is not injective and because multiple initial states map to the same outcome, the process is not invertible. Furthermore, because this operation is not trace-preserving for all density matrices, it does not qualify as deterministic physical evolution, distinguishing it from unitary evolution.

Also, if it is not the zero vector, what does the singlet evolve/ gets collapsed into?

It projects to the zero vector only within the subspace of coincidence detections, simply meaning that outcome is impossible.

Physically the state is not gone. Since the singlet consists of anticorrelated polarizations passing it through matched vertical polarizers guarantees that exactly one photon is absorbed and one is transmitted. The field does not decay to vacuum it decays to a single-photon state.

If we expand the Hilbert space to include the polarizers, the state evolves unitarily (some nuance here with we need to include the environment as well if we want to be accurate, with some doing heavy lifting) into an entangled superposition of one photon being absorbed and the other surviving:

$$|\psi_{final}\rangle \propto |\text{excited}\rangle_L \otimes |\text{photon}\rangle_R - |\text{photon}\rangle_L \otimes |\text{excited}\rangle_R$$

Now imagine some input state were both photons were absorbed. What happened to the state now?

It's gone. The state is undefined. If you only look at the transmitted state.

If we expand the Hilbert space to include the polarizers, the state evolves unitarily into something like, $|\text{no photon}\rangle \otimes |\text{excited polarizer}\rangle$

I hope that helps, I could also be incorrect and I've glossed over a few things. Hopefully some other member will jump in and correct me if that's the case.

 

[QuantumExplorer]

In QM we can look at two different update rules:
$$|\psi\rangle \mapsto U|\psi\rangle$$
with $U$ unitary. This evolution of the state vector is deterministic and preserves norm:
$$\|U|\psi\rangle\|=\||\psi\rangle\|$$
Alternatively, for say a measurement, each possible outcome is associated with an operator $A$ that salsifies $A^{\dagger}A \leq \mathbb{1}$, the outcome occurs with probability:
$$p=\|A|\psi\rangle\|^{2}$$
and if it happens, the (normalized) output state is:
$$|\psi'\rangle=\frac{A|\psi\rangle}{\sqrt{p}} \qquad\text{(only defined if }p>0\text{)} $$

The distinction between these updates is rooted in information conservation. Unitary evolution rotates the state vector without changing its length, ensuring reversibility. In contrast, say a measurement operator $A$ acts as a projection. It effectively deletes the components of the superposition that contradict the observed outcome and because multiple initial states could theoretically map to the same post-measurement state or zero, the mapping is not injective, and therefore not invertible. This loss of information renders the process irreversible and the outcome probabilistic. Getting $|\psi\rangle$ back from $|\psi'\rangle$ is impossible in this case.

As an aside and not to go too far off into lala land but there are also incomplete measurements that are not fully deterministic for the entire system but still preserve a probabilistic reversibility. (This is an invertible non-unitary A, see: https://qubit.guide/4.4-example-of-an-incomplete-measurement)


So for your example, if I have the singlet state and pass it through matched vertical polarizers, and we look specifically for coincidence counts, the probability is 0.

However, if I tell you no coincidence occurred, can you tell me with certainty that the input state was the singlet state? No. The input could have been the state $|HH\rangle$, which also yields zero coincidences. Even though the outcome appears deterministic, the state vector's evolution is not injective and because multiple initial states map to the same outcome, the process is not invertible. Furthermore, because this operation is not trace-preserving for all density matrices, it does not qualify as deterministic physical evolution, distinguishing it from unitary evolution.



It projects to the zero vector only within the subspace of coincidence detections, simply meaning that outcome is impossible.

Physically the state is not gone. Since the singlet consists of anticorrelated polarizations passing it through matched vertical polarizers guarantees that exactly one photon is absorbed and one is transmitted. The field does not decay to vacuum it decays to a single-photon state.

If we expand the Hilbert space to include the polarizers, the state evolves unitarily (some nuance here with we need to include the environment as well if we want to be accurate, with some doing heavy lifting) into an entangled superposition of one photon being absorbed and the other surviving:

$$|\psi_{final}\rangle \propto |\text{excited}\rangle_L \otimes |\text{photon}\rangle_R - |\text{photon}\rangle_L \otimes |\text{excited}\rangle_R$$

Now imagine some input state were both photons were absorbed. What happened to the state now?

It's gone. The state is undefined. If you only look at the transmitted state.

If we expand the Hilbert space to include the polarizers, the state evolves unitarily into something like, $|\text{no photon}\rangle \otimes |\text{excited polarizer}\rangle$

I hope that helps, I could also be incorrect and I've glossed over a few things. Hopefully some other member will jump in and correct me if that's the case.

I tried asking a new question at Stack Exchange to see if someone else chimes
https://quantumcomputing.stackexcha...ections/45938?noredirect=1#comment59732_45938

but I'm still at a loss in that the given definition of "invariance" seems contradictory. I agree that for non-unitary $A$ then the probability to get the singlet back (that's what I would call invariant!) is $|\det(A)|^2$. But when $A$ is singular, that chance is 0, so then it is NOT invariant.

 

[QuarkyMeson]

I mean, just so you know I agree with you. I don't think span preservation for quantum states is a good definition of invariance. The 0 vector isn't a quantum state. Your example shows that information has leaked out of the system into the environment, this is hardly what I would personally call physical invariance.

Mathematically, it is a valid definition though.