Simple example: Quantum error correction and Holography

One big puzzle for 21 century physics is Ads/CFT. It says that one quantum gravity in AdS (d+1 dimension ) is the same as quantum field theory (conformal field theory in d dimension )without Gravity.

Since I already publish two papers on it. There are thousands of papers discussing these topics already. But there are still many puzzles regarding this duality.
One of the puzzles comes from the radial commutativity. After I watched videos and look at papers written by Daniel Harlow which is an expert in this topics in MIT.
I think it is very interesting to discuss this duality by the toy model introduced by him.

Error Correction is a technique for people to protect the information via propagating.
For classical: if I want to send you one bit 1 or 0:
But in some channels? there are some chances the bit will be affected.
like 0 be 1 and 1 be 0?
How can we prevent this situation? we can use repetition code:
we repeat the single bit. i,e it means we send 000 (or 111) instead of sending only one bit 0 (or 1).
If you get our messages 101 then you know I want to send you 1!!!.
Idea:
$$ 0 \to 000 , 1 \to 111 $$ this is called repetition code. It can be shown that by using this method, we can decrease the chance of information being affected.

But in quantum mechanics, we have some troubles

No cloning theorem:
There is no unitary operator (Linear) which can copy any quantum state (unknown):

i.e there is no linear and unitary operator $$U$$ such that $$ U |v>=|v>\mathop{\otimes} |v> $$
 for any  \( |v> \) in Hilbert space.

proof:

Suppose this is possible: then Given |v> and |u>, we know  
$$<v|u>= <v| U^{\dagger}.U |u> =<v| \mathop{\otimes} <v| . |u> \mathop{\otimes}|u>=<v|u>^2$$ 
which leads to 
$$ <v|u>=0 \quad or \quad 1$$ this is means   |v> and |u> can not be arbitrary. Contradiction!!! 





Now we can imitate the quantum error correction code by encode 1 logical qubit into 3 physical qubit
Or define a linear map from 
\( \mathbb{C}^2 \to \mathbb{C}^8   \).
$$ a|0>+b|1>  \to  a|000>+b|111> $$
If the error correction happened. Define three error maps that may happened.

\( \epsilon_1:  a|000>+b|111> \to  a|100>+b|011>  \)

\( \epsilon_2:  a|000>+b|111> \to  a|010>+b|101>  \)

\( \epsilon_3:  a|000>+b|111> \to  a|001>+b|110>  \).
So now we have a good way solving this problem.
Suppose we have a state $$u = a|000>+b|111> $$. After it propagate, it may become  \( \epsilon_{i}u, i=1, 2, 3 \).
We build corresponding three operators : $$ Z_1 Z_2,  Z_1 Z_3,  Z_2 Z_3$$ which is pauli Z matrix act on each sites on  \( \epsilon_{i}u, i=1, 2, 3 \).


BUT NOW, we have $$ Z|0>=|0>, Z|1>=-|1>$$
So we have for example 
$$ Z_1Z_2 u = u $$
$$ Z_2Z_3 u = u $$
$$ Z_1Z_3 u = u $$

$$ Z_1Z_2   \epsilon_{1}u =  Z_1Z_2  (a|100>+b|011>)=-a|100>-b|011>=-u $$(since 1 and 2 site are different, one is zero, one is 1)
$$ Z_2Z_3  \epsilon_{1}u = Z_2Z_3  (a|100>+b|011>)=u $$ (since 2 and 3 site are all 0 or all 1)
$$ Z_1Z_3  \epsilon_{1}u = Z_1Z_3  (a|100>+b|011>)=-u $$ (since 1 and 3 site are different)

If we follow this logic to compute  \( \epsilon_{i}u, i= 2, 3 \). We can build a table.

SO now, if the error happened, we can know it by act this three operators, we can now which side of qubit flip.

If we know which side is flipped, we can just flip that qubit again by act pauli X matrix at that sides.

By Doing this, we can solve the questions. 

Now, what is the relationship between Ads/CFT to quantum error correction. It says that one quantum gravity in AdS (d+1 dimension ) is the same as quantum field theory (conformal field theory in d dimension )without Gravity.

Suppose we have an error which can eliminate quantum state.

how, so for example, if we have a|010>+b|111> which I want to send to my friend, if the error occurs,

we might have three possibilities.

How can we recover this state?

Can we produce a code, if one of qubits is loss, we can still recover the original quantum state.

This is three qutrix code from Daniel Harlow.

daniel harlow lectures

We have a logical qubit which has three energy levels called |0>, |1>, |2>.

Suppose we want to send $$|\phi> =a_0|0|>+a_1|1>+a_2|2>$$.

We now encode this logical qubit to three physical qubits.

which \( |\tilde{i}>\) is following 

The key step is we can find three unitary operators from 

$$U:  \mathbb{C}^{27}  \to  \mathbb{C}^{27} $$ such that :

\(U_{12} \) only acts on 1 and 2 qubits, \(U_{23} \) only acts on 23 qubits, \(U_{31} \) only acts on 13 qubits, 

What does this mean?

This means that if we loss the third qubit. We can use a small Quantum Computer \( U_{12}\) acting on the state product the highly entangle state \( X \)

So now, we get the situation improved. We can recover the state even with we loss one qubit.

What is this connection to Holography:

In lectures, we can view the logical qubit is in the bulk and the physical qubits are on the boundary.