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. U \cdot U^T = U^T \cdot U = U \cdot U^{-1} = I 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:

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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