# Quantum No-Deletion/No-Partial Erasure vs. Measurement

1. Oct 22, 2015

### Ruik

I'm currently working with several No-Go-Theorems in Quantum Mechanics for my master thesis and there are two, wich are confusing me: The No-Deletion-Theorem and the No-Partial Erasure-Theorem.
This is what I found out:
• The No-Deletion-Theorem shows that "there is no quantum deleting machine that can delete one onknown state against a copy in either a reversible or an irreversible manner."
It argues with the linearity of quantum operations.
• In the paper about the No-Partial-Erasure-Theorem it is said that "it is ompossible to erase quantum information, even partially and even by using irreversible means [...]"
• But some lines later: "Out theorem adds new insight into the integrity of quantum information, namely that we can erase complete information but not partial information."

So... I think it's important to look at the models used in these theorems. The No-Del-Theorem seems to focus on (unitary) transformations. Even though they talk about irreversible operations, measuring the qubits doesn't seem to be allowed there. That makes sense to me, because a unitary transformation is reversible, so the quantum information is retrieveable by applying the inverse operation.

The No-Part-Era-Theorem also is about reversible and irreversible operations, too, but measuring isn't discussed there either, as far as I can see.

My first question: What are these irreversible quantum operations, that can't be used to delete quantum information? As far as I know, there are two kinds of operations: unitary transformations (reversible) and measurements (changing the quantum state irreversibly).

My second question: Let |q>= a |0> + b |1> be an arbitrary qubit. I perform a measurement, wich brings |q> into the state either |0> or |1> and if it's |1> I flip it to |0> by rotating it by 90°.
Haven't I deleted the qubit? Each qubit will be mapped to |0>. It is irreversible, because I perform a measurement, but I don't see where the information could move to.

2. Oct 22, 2015

### jk22

For your second question since |q>=cos a|0>+sin a|1> and you measure you know the probabilities cos(a)^2 aso, you could almost reconstruct the qubit by a rotation of +\-tan a ? There are four possible qubits

3. Oct 22, 2015

### Strilanc

Measuring dissipates the information into the environment. That's not the same thing as deleting it.

For example, if you have three qubits in the GHZ state $\frac{1}{\sqrt{2}} \left| 000 \right\rangle + \frac{1}{\sqrt{2}} \left| 111 \right\rangle$ then deleting the third qubit should drop the remaining two down to a pure EPR state like $\frac{1}{\sqrt{2}} \left| 00 \right\rangle + \frac{1}{\sqrt{2}} \left| 11 \right\rangle$. Measuring the third qubit won't do that, it will drop the remaining two down to a mixed state. (Interestingly, if you measured along the correct axis then you can use the result to perform a phase correction that does leave you in an EPR state.)

4. Oct 22, 2015

### jk22

This deleting could only be performed by non square matrices.

5. Oct 23, 2015

### Ruik

Okay, thank you so far!

@jk22: I cannot reconstruct the qubit after a measurement, that qould be a big almost. You say you know the probabilities, but if I perform a measurement I only get an idea of the probability, because of the outcome of my measurement. And if I measure $|0\rangle$ it's not even sure, that $cos^2(a) > sin^2(a)$.

@Strilanc: Well yeay, if I perform a measurement, I get some information, but definetly not all of it. I get one bit of information, a qubit is capable of storing much more, even though it's not possible to get it. So am I not deleting most of the information?

Btw, your example differs from the situation the no deletion theorem assumes. They take two identical but not entangled qubits an an ancilla: $|q\rangle|q\rangle|A\rangle$
And according to them its impossible to transform it into $|q\rangle|0\rangle|A'\rangle$ without $|A'\rangle$ massively depending on $|q\rangle$ wich means, the information has been moved. They say this is the case for reversible ans irreversible operations. Any idea what irreversible operations they could mean?

6. Oct 23, 2015

### Strilanc

Well, the environment (and consequently you) become entangled with a qubit when it is measured. This prevents the alternative case from interacting with you anymore, due to decoherence. Whether or not the information is "really" gone depends on whether your favorite interpretation includes collapse.

It's misleading to think of measurement as "deleting" a qubit, because that leads directly to misunderstandings about how measurement affects entanglement. You can't kick a qubit out of a GHZ state by just measuring it, because doing so will kill the entanglement between the other two. A proper deletion operator would be $E = \begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}$. Hit a qubit with that, and it is forced into the Off state (deleting its value). Problem is, $E$ isn't unitary. You can kill all the amplitude in the system with it, e.g. |0> - |1> is annihilated by $E$. Not good.

I assume the irreversible operation they're talking about is collapse, or alternatively just the thermodynamic irreversibility of measurements that escape into the environment.