Unitary Transformations

Unitary transformations are like orthogonal transformations, except we’re working with a complex inner product space. We’ll focus on just the transformations that are unitary with respect to the inner product itself. That is, we ask that

$\displaystyle\langle w\vert v\rangle=\langle U(w)\vert U(v)\rangle=\langle w\rvert U^*U\lvert v\rangle$

and so we must have $U^*=U^{-1}$ just as we wrote for orthogonal transformations, but we have to use the adjoint that’s appropriate to our complex inner product. In this way, unitary and orthogonal transformations are related in a way similar to that in which Hermitian and symmetric forms are related.

Now, we’ve got this running analogy between endomorphisms on an inner product space and complex numbers. Taking the adjoint is like complex conjugation, so Hermitian transformations are like real numbers because they’re equal to their own adjoints. But here, we’re looking at transformations whose inverses are equal to their adjoints. What does this look like in terms of our analogy?

Well, we’ve noted that a transformation composed with its own adjoint is a sure way to get a positive-definite transformation $T^*T$ . This is analogous to the way that a complex number times its own conjugate is always nonnegative: $\bar{z}z\geq0$ . In fact, we use this to interpret $\bar{z}z$ as the squared-length of a the complex number $z$ . So what’s the analogue of the unitarity condition $U^*U=I_V$ ? That’s like asking for $\lvert z\rvert^2=\bar{z}z=1$ , and so $z$ must be a unit-length complex number. Unitary transformations are like the complex numbers on the unit circle.

July 28, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

9 Comments »

In Deutsche’s Problem, one is trying to compute $f(0)$ or $f(1)$ in 24 hours. Now if $f(x)$ is not invertible, why do we need a unitary transformation $U_f$ ?

E.g. we have $U_f: |x \rangle|y \rangle \to |x \rangle|y \oplus f(x) \rangle$ .

Comment by ramanujantao | July 28, 2009 | Reply
Because Deutsche is working on quantum computation, and all transformations in quantum mechanics are modeled with unitary operators on Hilbert spaces.

Comment by John Armstrong | July 28, 2009 | Reply
I see. The unitary transformation is essentially the “quantum computer.”

Comment by ramanujantao | July 28, 2009 | Reply
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