Distinguishing two preparations for same pure state leads to signalling

TL;DR

The paper proves that it is impossible to distinguish different preparation methods for the same pure quantum state, and doing so could enable signaling, highlighting fundamental limits in quantum state preparation and communication.

Contribution

It introduces a novel impossibility result linking indistinguishability of preparations to no-signaling, extending to bipartite entangled states.

Findings

Distinguishing preparations for a pure state is impossible.

Distinguishing preparations can lead to signaling.

Extension to bipartite states shows similar impossibility.

Abstract

Pure state of a physical system can be prepared in an infinite number of ways. Here, we prove that given a pure state of a quantum system it is impossible to distinguish two preparation procedures. Further, we show that if we can distinguish two preparation procedures for the same pure state then that can lead to signalling. This impossibility result is different than the no measurement without disturbance and the no-cloning. Extending this result for a pure bipartite entangled state entails that the impossibility of distinguishing two preparation procedures for a mixed state follows from the impossibility of distinguishing two preparations for a pure bipartite state.

Equations

$$U|0⟩|A⟩$$

$$V|+⟩|A⟩$$

$$1=⟨A_{U_0}||A_{V_{+}}⟩.$$

$$|0⟩|A⟩$$

$$|+⟩|A⟩$$

$$|{\Psi }^{-}⟩=\frac{1}{\sqrt{2}}(|0⟩|1⟩-|1⟩|0⟩).$$

$$|{\Psi }^{-}⟩=U(\alpha ,\beta )\otimes U(\alpha ,\beta )|{\Psi }^{-}⟩=\frac{1}{\sqrt{2}}(|\psi ⟩|\bar{\psi }⟩-|\bar{\psi }⟩|\psi ⟩),$$

$$U^{†}(\alpha ,\beta )\otimes I|{\Psi }^{-}⟩$$

$$V^{†}(\alpha ,\beta )\otimes I|{\Psi }^{-}⟩$$

$$U|0⟩|A⟩$$

$$U|1⟩|A⟩$$

$$\frac{1}{\sqrt{2}}(|0⟩U|1⟩|A⟩-|1⟩U|0⟩|A⟩)$$

$$\frac{1}{\sqrt{2}}(|0⟩|\bar{\psi }⟩|A_{U_1}⟩-|1⟩|\psi ⟩|A_{U_0}⟩),$$

$$\frac{1}{\sqrt{2}}(|+⟩V|-⟩|A⟩-|-⟩V|+⟩|A⟩)$$

$$\frac{1}{\sqrt{2}}(|0⟩|\bar{\psi }⟩|A_{V_{-}}⟩-|1⟩|\psi ⟩|A_{V_{+}}⟩).$$

$${\rho }_B^0=\frac{1}{2}[|\bar{\psi }⟩⟨\bar{\psi }|\otimes |A_{U_1}⟩⟨A_{U_1}|+|\psi ⟩⟨\psi |\otimes |A_{U_0}⟩⟨A_{U_0}|],$$

$${\rho }_B^{+}=\frac{1}{2}[|\bar{\psi }⟩⟨\bar{\psi }|\otimes |A_{V_{-}}⟩⟨A_{V_{-}}|+|\psi ⟩⟨\psi |\otimes |A_{V_{+}}⟩⟨A_{V_{+}}|],$$

Full text

In quantum mechanics, the state of a physical system, be it pure or mixed, is supposed to capture the complete description of a system. It is known that if we describe the system by a mixed state, then the same mixed state can be prepared in an infinite number of ways by probabilistic mixing of different decompositions of the same density matrix. Once the state is prepared, it is impossible in principle to distinguish two (or more) preparation procedures for a mixed state. In fact, if we can distinguish two preparations for a mixed state, then we could have signalling and we could violate the Second law of thermodynamics peres .

If a physical system is in a pure state, that can also be prepared in an infinite number of ways. Since a Hilbert space can have an infinite number of orthonormal basis sets, we can expand a pure state using any orthonormal basis. Consider two observables $A$ and $B$ with eigenbasis sets $\{|{a_{n}}\rangle\}$ and $\{|{b_{n}}\rangle\}$, respectively. Then, we know that we can express the pure state using the eigenbasis of the observables $A$, i.e., $|{\psi}\rangle=\sum_{n}\alpha_{n}|{a_{n}}\rangle$ or we could also expand the pure state using the eigenbasis of the observables $B$, i.e., $|{\psi}\rangle=\sum_{n}\beta_{n}|{b_{n}}\rangle$. Each of these possible expansions represents one possible preparation procedure. In quantum mechanics, we can prepare a pure state starting from a fiducial pure state either by a unitary transformation, or by a general quantum operation. Since this has to be a physical operation, and any physical operation can be realized by a unitary transformation of the fiducial state along with an ancillary state, without loss of generality, we consider here only unitary operation as a physically realizable process to prepare a pure state.

To convey the main result, we consider a quantum system in a two-dimensional Hilbert space and one can generalize our results to higher dimensional systems. For example, consider a qubit in a state $|{\psi}\rangle=\alpha|{0}\rangle+\beta|{1}\rangle$. This can be prepared starting from an initial state $|{0}\rangle$ (say along up-$z$-axis of a spin-half particle) and by applying a unitary $U(\alpha,\beta)$, i.e., $|{\psi}\rangle=U(\alpha,\beta)|{0}\rangle$. The same state can also be prepared starting from an initial state $|{+}\rangle$ (say up-$x$-axis of a spin-half particle) and by applying a different unitary $V(\alpha,\beta)$, i.e., $|{\psi}\rangle=V(\alpha,\beta)|{+}\rangle$. The question we address is whether it is possible to distinguish these two preparation procedures by any physical operation. To our surprise, we discovered that the impossibility of distinguishing two preparation procedures for a pure state has never been proved in the literature, not to mention about how that can lead to signaling. Here, we show that it is impossible to distinguish two (or more) preparations for a pure state. Then, we prove that if we can distinguish two different preparations for a pure state, then one can have signaling. Also, we argue that the new impossibility result is different than the other no-go theorems such as the impossibility of measuring without disturbance park and the no-cloning in quantum information wz ; dd . Towards the end, we prove that the impossibility of distinguishing two preparation procedures for a mixed state follows from the impossibility of distinguishing two preparations for a pure bipartite state.

Suppose that there is a machine which can distinguish two preparation procedures for a pure state. If the machine respects quantum mechanics, then that has to be represented by a physical operation. We can always realize a physical operation as a unitary evolution on a larger Hilbert space. Imagine that the physical operation is a unitary operation on the system and the machine state. Now, assume that the machine somehow knows the preparation procedure and the final state of the machine changes according to the preparation procedure. Thus, the unitary transformation that may distinguish two preparations for a qubit state is given by

[TABLE]

where $|{A}\rangle$ is the initial state of the machine, $|{A_{U_{0}}}\rangle$ is the final state of the machine if the pure state is prepared via $U|{0}\rangle$ and $|{A_{V_{+}}}\rangle$ is the final state of the machine if the pure state is prepared via $V|{+}\rangle$. The ‘preparation-distinguishing’ machine, if it exists, then that will change the final state of the machine according to the preparation procedure and leaves the input state unchanged. Note that we need the states $|{A_{U_{0}}}\rangle$ and $|{A_{V_{+}}}\rangle$ to be different, so that we can obtain information about the preparation procedures of the pure state. However, if this is to hold, then we have

[TABLE]

This implies that $|{A_{U_{0}}}\rangle$ and $|{A_{V_{+}}}\rangle$ can never be different and hence there is no way to distinguish two preparation procedures for a pure state.

Note that the machine that we have defined above is distinct one compared to the machine that is supposed to measure two non-orthogonal quantum states without disturbance In this case, the transformation is defined as

[TABLE]

That this process is also impossible follows from the unitarity, because we can never be able to satisfy $1=\langle{A_{0}}||{A_{+}}\rangle$. This is paraphrased by saying that ‘it is impossible to distinguish two non-orthogonal states without disturbance’. However, note that these two machines are completely different. This is because there is no way that we can go from Eq(1) to Eq(2) as the unitaries $U$ and $V$ in general will not commute with the global operation that realizes the process given in Eq(1). Therefore, the hypothetical machine that can distinguishing two preparation procedures for a pure state is fundamentally different than the machine that is supposed to distinguish two non-orthogonal quantum states without disturbance. Therefore, this impossibility is a new result independent of the earlier one. We can also argue that it is independent of the no-cloning theorem wz ; dd . First, note that if we know the complete preparation procedure, then we know the state of a qubit. But the converse is not true, i.e., knowing the state of a qubit is not same as knowing the preparation procedure. If we know the state of a qubit, then we can clone it, whereas here, even if we know the state we cannot distinguishing two preparation procedures. This shows that the present no-go theorem is different from the other no-go theorems such as the no-cloning wz ; dd and the no-deleting theorems pati .

Distinguishing two preparations for pure state and signalling: Here, we will show that distinguishing two different preparations for a pure state can actually lead to signalling. Imagine that Alice and Bob share an entangled Einstein-Podolsky-Rosen (EPR) pair as given by

[TABLE]

The EPR state satisfies the property

[TABLE]

where $|{\psi}\rangle=\alpha|{0}\rangle+\beta|{1}\rangle=U(\alpha,\beta)|{0}\rangle$ and $|{{\bar{\psi}}}\rangle=\alpha^{*}|{1}\rangle-\beta^{*}|{0}\rangle=U(\alpha,\beta)|{1}\rangle$. This invariance property of singlet is equivalent to

[TABLE]

Physically, this means that if Alice applies $U^{\dagger}(\alpha,\beta)$ on her particle this is equivalent to applying $U(\alpha,\beta)$ on Bob’s particle. Similarly, the invariance property for the singlet implies that we have

[TABLE]

with the notion that $|{\psi}\rangle=V(\alpha,\beta)|{+}\rangle$ and $|{{\bar{\psi}}}\rangle=V(\alpha,\beta)|{-}\rangle$. Now, let us encode one classical bit in Alice’s action, i.e., if she receives [math], then she applies $U^{\dagger}(\alpha,\beta)$ on her particle and if she receives $1$, then she applies $V^{\dagger}(\alpha,\beta)$ on her particle. These two choices by Alice allow us to have the possibility of two different preparations at Bob’s end. Now assume that Bob has a hypothetical machine which can distinguish two preparation procedures for a pure state. Bob attaches the machine and allows the transformation as given by

[TABLE]

Then, depending on the two choices of preparations of a pure state, we have

[TABLE]

Now, the two preparation procedures gives two different density matrices at Bob’s end. These are given by

[TABLE]

where $\rho_{B}^{0}$ is the result of one preparation procedure and $\rho_{B}^{+}$ is the result of other preparation procedure. Since these two density matrices are different, Bob can infer Alice’s action, thus revealing one bit of information without any communication from Alice. This would lead to signaling. Therefore, from the no-signaling, we can argue that it is impossible to distinguish two preparation procedures for a pure state.

Impossibility of distinguishing two preparations for bipartite states: Now, we ask can the impossibility of distinguishing two preparations for a pure bipartite state lead to the impossibility of distinguishing preparations for a mixed state? In quantum mechanics, a density matrix can have infinite number of decompositions (proper mixtures) and it is impossible to distinguish two preparation procedures. Also, we know that a mixture (improper) occurs when we trace out one of the subsystem of an entangled state. In what follows, we show that the impossibility to distinguish two preparations for a pure bipartite entangled state indeed implies the impossibility of distinguishing two preparation procedures for a mixed state.

First, note that a pure bipartite entangled state can also be prepared in an infinite number of ways. Consider two preparation procedures of a pure bipartite state $|{\Psi}\rangle_{AB}$ which are expressed as $\sum_{nm}C_{nm}|{\psi_{n}}\rangle|{\phi_{m}}\rangle$ and $\sum_{\mu\nu}\alpha_{\mu\nu}|{a_{\mu}}\rangle|{b_{\nu}}\rangle$. Extending our earlier result to a pure bipartite state, we can show that it is impossible to distinguish two different preparations. Now, these two preparation procedures for a pure state will result in two possible preparations for the density matrix of either subsystem. For example, if we trace out the second subsystem, we will have density matrix $\rho_{A}=\sum_{m}|{{\tilde{\psi}}_{m}}\rangle\langle{{\tilde{\psi}}_{m}}|$, where $|{{\tilde{\psi}}_{m}}\rangle=\sum_{n}C_{nm}|{\psi_{n}}\rangle$ are unnormalized and non-orthogonal states with $\sum_{m}||{\tilde{\psi}}_{m}||^{2}=1$. Similarly, for the other preparation procedure, if we trace out the second subsystem, then the density matrix $\rho_{A}=\sum_{\nu}|{{\tilde{a}}_{\nu}}\rangle\langle{{\tilde{a}}_{\nu}}|$, where $|{{\tilde{a}}_{\nu}}\rangle=\sum_{\mu}\alpha_{\mu\nu}|{a_{\mu}}\rangle$ are unnormalized and non-orthogonal states with $\sum_{\nu}||{\tilde{a}}_{\nu}||^{2}=1$.

Now, suppose that there is a physical operation that can perfectly distinguish two preparations for the same mixed state. This means that the ‘preparation-distinguishing’ machine can result in two different states of the system and the machine as $\rho_{AE}^{(1)}$ and $\rho_{AE}^{(2)}$ corresponding to two different preparation procedures, respectively, with $D(\rho_{AE}^{(1)},\rho_{AE}^{(2)})=1$. Here $D$ is a measure of distinguishing two different preparations which are labeled as $`1^{\prime}$ and $`2^{\prime}$. This physical operation can also be realized on a purified Hilbert space that results in two possible states $|{\Psi_{ABE}^{(1)}}\rangle$ and $|{\Psi_{ABE}^{(2)}}\rangle$ according to two different preparation procedures. Note that the distinguishabilty measure decreases under partial tracing, therefore, we have $D(\rho_{AE}^{(1)},\rho_{AE}^{(2)})\leq D(\Psi_{ABE}^{(1)},\Psi_{ABE}^{(2)})$. Since we have assumed that $D(\rho_{AE}^{(1)},\rho_{AE}^{(2)})=1$, this then implies that $D(\Psi_{ABE}^{(1)},\Psi_{ABE}^{(2)})\geq 1$. If the distinguishabilty measure satisfies $0\leq D\leq 1$, then it must be true that $D(\Psi_{ABE}^{(1)},\Psi_{ABE}^{(2)})=1$, i.e., we can distinguish two preparations for a pure bipartite entangled state perfectly. But we know that we cannot distinguish two different preparations for the same pure entangled state and hence it is impossible to distinguish two preparations for the same mixed state.

Conclusions: In quantum theory, the preparation of a physical system and the measurement procedure play fundamental role. Measurement process though always entails an outcome which may be random, an experimentalist must be able to reproduce the preparation and measurement procedures. A reproducible preparation of a physical system is represented by a pure state (in the case of closed system) or by a density operator (in the case of open system). There are infinite number of ways in which a given pure state can be prepared and hence there is an infinite number of pasts associated to a present pure state of a physical system. Our results shows that once the state is prepared in a pure state, then there is no way to reveal its preparation procedure. We have also shown that the violation of the impossibility of distinguishing two different preparations for a pure state can lead to signalling. Moreover, we have argued that our result is independent of the no measurement without disturbance and the no-cloning. We have also proved that the impossibility of distinguishing two preparations for the same mixed state follows from a more fundamental result that it is impossible to distinguish two different preparations for the same pure bipartite state. The new impossibility result, hitherto unnoticed, has a different status compared to other no-go theorems such as the no-cloning and the no-deleting theorems in quantum information, and opens up several questions of fundamental importance in quantum theory.

Note: I thank Ujjwal Sen and Chirag Srivastva for useful discussions. For another perspective on distinguishing and signalling readers can see today’s paper in the arXiv by Srivastva et al.