# Tag Info

Accepted

### What's so bad about preparing magic states by measuring Clifford gates?

what is so bad about fault tolerantly preparing magic states by measuring Clifford gates? There's nothing bad about it. See "Very low overhead fault-tolerant magic state preparation using ...
• 40.1k

### GHZ Definition - Does it all have to be $|000\ldots\rangle$ or $|111\ldots\rangle$ or can it be $|010\ldots\rangle$ or $|101\ldots\rangle$

TL;DR: No, these are not GHZ states. They are cat states. GHZ state The GHZ state is defined as the state $$\frac{|000\rangle+|111\rangle}{\sqrt{2}}.\tag1$$ Generalized GHZ ...
• 23.6k
Accepted

• 60.3k
1 vote

### Transformation between the canonical basis and the degraded Bell basis

I understood why I did not get the correct result. It's simply because in the book I am reading they just used, at the begining, a lightly different degraded Bell basis. More precisally this one: \...
• 193
1 vote
Accepted

### Polar decomposition of $\sqrt{\sqrt{\rho} \sigma \sqrt{\rho}}$

Note that $\sqrt\rho\sigma\sqrt\rho$ is already positive semidefinite$^1$, so defining $A=\sqrt\rho\sigma\sqrt\rho$ leads to $|A|=A$. Not interesting. Instead, let's define $A:=\sqrt\sigma\sqrt\rho$. ...
• 23.6k
1 vote

### Effecient creation of a sudoku/permutatation state

Expanding on a comment above, the following might work with a little bit of error and would be polynomial in $n$, but would probably be a brutal polynomial with a very deep circuit. Initially prepare ...
• 13.2k
1 vote

### How to calculate guessing probability for quantum key distribution?

It's not easy. I assume that you are working here in the device-independent scenario. The first thing you need is to understand how the noise affects the value of the Bell inequality; this value ...
• 2,853
1 vote

### What exactly is the computational or standard basis?

You are absolutely right in your statement that there is a freedom in the choice of basis, for instance when looking at an expectation value of an observable $O$ on a state $\lvert \psi \rangle$, you ...

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