Solar chimney

Quantum states are fragile in one sense and also robust in another sense. Quantum theory tells us that given a single quantum it is impossible to determine it. One needs infinite number of identically prepared quantum states (copies) to know a state exactly. This has remarkable consequences in quantum information theory. One is the familiar no-cloning theorem for a single quantum. Similar to the quantum no-cloning theorem, there is another no-go theorem in quantum information theory which is called as the no-deleting theorem

Quantum deleting

Suppose that there are two copies of an unknown quantum state. A pertinent question in this context is to ask if it is possible to delete one copy from two copies using quantum mechanical operations? It turns out that we cannot. The no-deleting theorem is a consequence of linearity of quantum mechanics. Like the no-cloning theorem this has important implications in quantum computing, quantum information theory and quantum mechanics in general.

The process of quantum deleting takes two copies of an arbitrary, unknown quantum state at the input port and outputs a blank state along with the original. Mathematically, this can be described by:

$U|\psi \rangle _{A}|\psi \rangle _{B}|A\rangle _{C}=|\psi \rangle _{A}|0\rangle _{B}|A'\rangle _{C}$,

where $U$ is the deleting operation which is not necessarily unitary (but a linear operator), $|\psi \rangle _{A}$ is the unknown quantum state, $|0\rangle _{B}$ is the blank state, $|A\rangle _{C}$ is the initial state of the deleting machine and $|A'\rangle _{C}$ is the final state of the machine.

It may be noted that classical bits can be copied and deleted, so also qubits in orthogonal states. For example, if we have two identical qubits $|00\rangle$ and $|11\rangle$ then we can transform to $|00\rangle$ and $|10\rangle$. In this case we have deleted the second copy. However, it follows from linearity of quantum theory that there is no $U$ that can perform the deleting operation for any arbitrary state $|\psi \rangle$.

Theorem

Let $|\psi \rangle$ is an unknown quantum state in some Hilbert space (and other states have their usual meaning). Then, there is no linear isometric transformation such that $|\psi \rangle _{A}|\psi \rangle _{B}|A\rangle _{C}\rightarrow |\psi \rangle _{A}|0\rangle _{B}|A'\rangle _{C}$, with the final state of the ancilla being independent of $|\psi \rangle$.

Proof

The theorem holds for quantum states in any Hilbert space dimension. For simplicity, let us consider the deleting transformation for two identical qubits. If two qubits are in orthogonal states then we have

$|0\rangle _{A}|0\rangle _{B}|A\rangle _{C}\rightarrow |0\rangle _{A}|0\rangle _{B}|A_{0}\rangle _{C}$,

$|1\rangle _{A}|1\rangle _{B}|A\rangle _{C}\rightarrow |1\rangle _{A}|0\rangle _{B}|A_{1}\rangle _{C}$.

Let $|\psi \rangle =\alpha |0\rangle +\beta |1\rangle$ be the state of an unknown qubit. If we have two copies of an unknown qubit, then by linearity of the deleting trtansformation we have

$|\psi \rangle _{A}|\psi \rangle _{B}|A\rangle _{C}=[\alpha ^{2}|0\rangle _{A}|0\rangle _{B}+\beta ^{2}|1\rangle _{A}|1\rangle _{B}+\alpha \beta (|0\rangle _{A}|1\rangle _{B}+|1\rangle _{A}|0\rangle _{B})]|A\rangle _{C}$

$\rightarrow \alpha ^{2}|0\rangle _{A}|0\rangle _{B}|A_{0}\rangle _{C}+\beta ^{2}|1\rangle _{A}|0\rangle _{B}|A_{1}\rangle _{C}+{\sqrt {2}}\alpha \beta |\Phi \rangle _{ABC}.$

In the above expression, the following transformation has been used: $1/{\sqrt {2}}(|0\rangle _{A}|1\rangle _{B}+|1\rangle _{A}|0\rangle _{B})|A\rangle _{C}\rightarrow |\Phi \rangle _{ABC}$.

However, if we are able to delete a copy, then at the output port of the deleting machine the combined state should be

$|\psi \rangle _{A}|0\rangle _{B}|A'\rangle _{C}=(\alpha |0\rangle _{A}|0\rangle _{B}+\beta |1\rangle _{A}|0\rangle _{B})|A'\rangle _{C}$.

In general, these states are not identical and hence we can say that the machine fails to delete a copy. If we require that the final output states are same, then we will see that there is only one option:

$|\Phi \rangle =1/{\sqrt {2}}(0\rangle _{A}|0\rangle _{B}|A_{1}\rangle _{C}+|1\rangle _{A}|0\rangle _{B}|A_{0}\rangle _{C})$, and

$|A'\rangle =\alpha |A_{0}\rangle _{C}+\beta |A_{1}\rangle _{C}$.

Since final state of the ancilla is normalized for all values of $\alpha ,\beta$ it must be true that $|A_{0}\rangle$ and $|A_{1}\rangle$ are orthogonal. This means that the quantum information is simply in the final state of the ancilla. One can always obtain the unknown state from the final state of the ancilla using local operation on the ancilla Hilbert space. Thus, linearity of quantum theory does not allow an unknown quantum state to be deleted perfectly.

Though perfect quantum deleting is not possible, it is possible to perform imperfect deleting using deterministic operations and perfect deletion of linearly independent states using probabilistic operations.

Consequence

• If it will be possible to delete an unknown quantum state then using two pairs of EPR state we can send signal faster than light. Thus, the no-deleting theorem is in consistent with the no-signalling condition.

• The no-cloning and the no-deleting theorems point to conservation of quantum information.

• A stronger version of the no-cloning theorem and the no-deleting theorem provide permanence to quantum information. To create a copy one must import the information from some part of the universe and to delete one needs to export it to other part of the universe where it will continue to exists.

References

• W.K. Wootters and W.H. Zurek, A Single Quantum Cannot be Cloned, Nature 299 (1982), 802.
• D. Dieks, Communication by EPR devices, Physics Letters A, vol. 92(6) (1982), 271.
• A. K. Pati and S. L. Braunstein, Impossibility of Deleting an Unknown Quantum State, Nature 404 (2000), 104.