# Lecture 004 - Inverse and Reverse

## H.A.T. in Matrix Representation

You can merge probability and amplitude on the same layer of the tree.

You can summarize a computation by transitional probability (amplitude) matrix. (where columns labeled by input binary string, output labeled by output binary string). Function composition is just matrix multiplication. With amplitude, instead of $(\forall j)(\sum_{i} A_{ij} = 1)$, you have $(\forall j)(\sum_{i} \|A_{ij}\|^2 = 1)$ where $A_{ij}$ is a complex number.

With $n$ variables, there are $2^n$ states to track. Matrix size is $2^n \times 2^n$.

So what is the reverse and inverse of operations?

## Reverse and Inverse

Reverse: transpose of matrix that is a valid computation Inverse: exact undo

$\mathcal{U}$ is physically possible $\iff \mathcal{U}^{-1} = \iff \mathcal{U}^{\dagger}$. If so, $\mathcal{U}$ is unitrary (keep vector length).

Note that quantum operations do not need to be unitary, but quantum gates are unitary. Measurements are arguably the most obvious example of non-unitary transitions being a fundamental component of algorithms. (from Stackoverflow)

Unitrary matrix maps quantum states to quantum states. (Therefore $(\forall j)(\sum_{i} \|A_{ij}\|^2 = 1)$ holds)

### Classical

Classical: all columns only has one entry that is one, other entries are all zeros.

Reverse: might not exists, unless bijection Inverse: might not exists, unless bijection

Property: inverse is reverse

For non-probabilistic function that happens to be bijection, the reverse operation is the same as inverse operation.

### Probabilistic

Reverse: might not exist for certain instructions (Time-Reversible) Inverse: can't exist for all probabilistic instructions

You could find the inverse of a matrix, but that does not represent a valid inverse instruction.

### Quantum (Real Physics)

Reverse ($\dagger$): reverse arrow and replace $i$ to $-i$. Inverse ($^{-1}$): $(HAT)^{-1} = (HAT)$

Reverse ($M^\dagger$), Transpose ($M^T$), and Inverse ($M^{-1}$) are totally different concept.

Property: inverse is reverse

Since inverse is reverse, by definion, every valid physical operation is a unitary matrix, and every unitary matrix is a valid physical operation. $ where we have $U^{T} = U^{\dagger}$ given $U$ is a real matrix (if it contain imaginary entries, you need to exchange the real and imaginary part in addition to taking the inverse)

(HAT)^{-1} = (HAT) = (HAT)^{\dagger}

## $\sum_i a_i^2 = 1$

Assuming we somehow got some amplitudes $\sum_i a_i^2 = 1$ after some operation $U$, then:

• $\sum_i a_i^4 = 1$: this is the amplitude after applying $U^\dagger \cdot U$, which preserves the amplitude.

• $\sum_i (a_i^2w_i^2)^2 = \sum_i a_i^4w_i^4 = \sum_i a_i^4 = 1$: because this is the amplitude after applying $U^\dagger \cdot V^\dagger \cdot V \cdot \cdot U$

I don't understand the above explanation, but here is a linear algebra explanation.

Say that we have a amplitude vector $v$, and a quantum operation $U$. Observe that $v^T \cdot v = 1$ captures the property that each entry's square sum up to one (therefore $v$ is a valid quantum state).

We want to prove that the output amplitude $U \cdot v$'s square also sum up to one. It is equivalent to prove that $(U \cdot v)^T \cdot (U \cdot v)$ sum up to one.

\begin{align*} 1 =& (U \cdot v)^T \cdot (U \cdot v)\\ \iff 1 =& (U \cdot v)^\dagger \cdot (U \cdot v) \tag{by transpose is inverse for reals} \\ \iff 1 =& (v^\dagger \cdot U^\dagger) \cdot (U \cdot v) \tag{by property of inverse}\\ \iff 1 =& v^\dagger \cdot U^\dagger \cdot U \cdot v\\ \iff 1 =& v^\dagger \cdot v \tag{by property of inverse}\\ \iff 1 =& v^T \cdot v \tag{by transpose is inverse for reals}\\ \end{align*}

The above proof will only work for real numbers. Since for real numbers $M^T = M^\dagger$. But it is not true if $M$ contains complex entries.

Inverse $\square^{-1}$ is equal to reverse $\square^\dagger$ only if the matrix is unitary matrix. But inverse $\square^{-1}$ is equal to transpose $\square^T$ only if the matrix is orthogonal. Orthogonal matrices preserve the sum-of-squares-of-entries when they operate on vectors.

The above proof is only valid if you replace $\square^T$ with $\square^\dagger$. A unitary matrix $U$ does not satisfy $\square^T = U^{-1}$, but rather $U^\dagger = U^{-1}$. Also vector $v$ is only a quantum state iff the sum of the squares of the absolute values of its entries is $1$, that is $v^\dagger \cdot v = 1$ (only relevent when $v$ has imaginary entry)

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