A hermitian form H(,) on a complex vector space satisfies:

  • H(x,y) is linear in x for all y
  • H(x,y) = H(y,x)* for all vectors x and y

Given such a from on a complex vector space, it defines a natural way to associate the space with its dual. Knowing this, we call the dual map to a map on the space its hermitian adjoint. If a map is equal to its hermitian adjoint, it is said to be hermitian.

In Quantum Mechanics, the hermitian adjoint of an operator A is denoted A+, and is defined to be the operator that satisfies
    <x|A|y> = <y|A+|x>*
for all states labeled by x and y. By taking these labels to be discrete, the matrix Aij = <i|A|j> can be said to be hermitian if its transpose is equal to its complex conjugate.

Hermitian operators are said to correspond to observables (that is, their eigenvalues are the quantities that are measured by experiment), which is awfully handy since their eigenvalues are all real.

A matrix is said to be hermitian if it is equal to its adjoint (or hermitian conjugate), which is in other words in compliance with servus' writup above.

A hermitian operator is one that satisfies < Hx|y > = < x|Hy >, where x and y are real or complex vectors or functions and H is a linear operator and < x|y > denotes the inner product of x and y.

Some properties: Suppose we know that Hx = λx where &lambda is an eigenvalue of the hermitian operator H. Then we can prove that &lambda must be real. Consider:

< Hx|x > = < x|Hx >, therefore < λx|x > = < x|λx > which by the linearity or the inner product reduces to: &lambda*< x|x > = &lambda< x|x > (where * denotes the complex conjugate). Since cannot be zero by the non-negativity of the inner product (here x represents a magnitude which is by definition non-negative), &lambda* = &lambda therefore &lambda must be real.

Given that x and y are eigenvectors of the hermitian operator H, and &lambda and ɸ are the eigenvalues of H to x and y respectively (note also that &lambda ≠ ɸ), then we can prove that x and y are orthogonal.

Consider that Hx = λx and Hy = ɸy. Now, < Hx|y > = < x|Hy > which means that < λx|y > = < x|ɸy > so by the linearity of the inner product, &lambda< x|y > = ɸ< x|y > . Since &lambda ≠ ɸ, < x|y > = 0, ergo x is orthogonal to y.

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