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In quantum mechanics and quantum computing, quantum particles evolve in a unitary manner. That is to say at any point in time the particle/system (represented as a vector) has a magnitude of 1, meaning that all the different probabilities of its states sum to 100%.

This is all fine and well, but why is it that only matrices that are their own inverse (i.e unitary matrices) satisfy this property? Do they preserve the magnitude all vectors? or just unit vectors?

EDIT: My bad unitary matrices are the inverse of their own Hermitian conjugate.

Ozaner Hansha
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  • A unitary matrix is one whose conjugate transpose is its inverse, i.e. $U$ is unitary iff $UU=I$ where $U$ is $U$ but with every element conjugated and flipped along the main diagonal. – Eben Kadile Jul 14 '18 at 01:45
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    From an intuitive standpoint, unitary matrices represent rotations, which should not change the length of any vector. – Adrian Keister Jul 14 '18 at 02:04

2 Answers2

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No, a unitary matrix is not its own inverse. It is the inverse of its Hermitian conjugate.

Using the polarization identity, a matrix preserves magnitudes of vectors (all vectors, not just unit vectors) if and only if it preserves the inner product, and that is true if and only if the matrix is unitary:

$$ \langle U x, U y \rangle = \langle x, U^* U y \rangle = \langle x, y \rangle \ \text{for all} \; x,y \; \text{iff}\ \ U^* U = I$$

Robert Lewis
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Robert Israel
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  • I think you might want to add "$= \langle x, x \rangle = \Vert x \Vert^2$" or similar on the right; otherwise what you have written in your equation holds whether $U^\dagger U = I$ or not. Cheers! from an admirer! – Robert Lewis Jul 14 '18 at 01:49
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Unitary matrices are invariant under the 2 norm. You can show this like the following.

$$\| Qx\|_{2}^{2} = \|x\|_{2}$$

$$ \| Qx \|_{2}^{2} = (Qx)^{*}Qx $$ $$ =x^{*}Q^{*}Qx $$ $$ =x^{*}\underbrace{Q^{-1}Q}_{I}x $$ $$ =x^{*}x $$ $$ =\|x\|_{2}^{2} $$