410
7 Variational Principles
(T x, x). Now that (T x, x) is equal to the conjugate number of itself, so (T x, x) is a
real number.
The sufficiency. The following identity can be proved
4(T x, y) =[T (x + y), x + y] − [T (x − y), x − y]
+ i[T (x + iy), x + iy] − [T (x − iy), x − iy]
According to the condition of the theorem, all inner products on the right side are
real numbers. Transposing x and y, we obtain
4(T y, x) =[T (y + x), y + x] − [T (y − x), y − x]
+ i[T (y + i x), y + i x] − [T (y − i x), y − i x]
According to the basic property of inner product, there is
[T (y − x), y − x] = [T (x − y), x − y]
[T (y + i x), y + i x] = [i T (i x − y), i(x − iy)] = [T (x − iy), x − iy]
[T (y − i x), y − i x] = [−i T (x + iy), −i(x + iy)] = [T (x + iy), x + iy]
Thus it can be concluded that (T x, y) = (T y, x), or according to the symmetry
of inner product (T x, y) = (x, T y), namely the operator T is symmetric. Quod erat
demonstrandum.
If T is a self-conjugate operator, then
(T x, y) = (x, T
∗ y) = (x, T y) = (T y, x)
It follows that the self-conjugate operator must be the symmetric operator, but the
symmetric operator is not necessarily the self-conjugate operator. thus, by theorem
7.4.4 the following corollary can be obtained:
Corollary 7.4.1 The necessary and sufficient conditions that the operator T is the
self-conjugate operator are: The inner product (T x, x) is a real number.
Theorem 7.4.5 Let T 1 and T 2 be both self-conjugate operators, then the necessary
and sufficient conditions that T 1 T 2 is also a self-conjugate operator are T 1 T 2 = T 2 T 1 .
Proof The necessity. If T 1 T 2 is a self-conjugate operator, then for any x, y ∈ H ,
there is
(T 1 T 2 x, y) = (x, T 1 T 2 y)
Since T 1 and T 2 are both self-conjugate operators, and they are the bounded linear
operator from the Hilbert space H to the Hilbert space H , there is
(T 1 T 2 x, y) = (T 2 x, T 1 y) = (x, T 2 T 1 y)
7 Variational Principles
(T x, x). Now that (T x, x) is equal to the conjugate number of itself, so (T x, x) is a
real number.
The sufficiency. The following identity can be proved
4(T x, y) =[T (x + y), x + y] − [T (x − y), x − y]
+ i[T (x + iy), x + iy] − [T (x − iy), x − iy]
According to the condition of the theorem, all inner products on the right side are
real numbers. Transposing x and y, we obtain
4(T y, x) =[T (y + x), y + x] − [T (y − x), y − x]
+ i[T (y + i x), y + i x] − [T (y − i x), y − i x]
According to the basic property of inner product, there is
[T (y − x), y − x] = [T (x − y), x − y]
[T (y + i x), y + i x] = [i T (i x − y), i(x − iy)] = [T (x − iy), x − iy]
[T (y − i x), y − i x] = [−i T (x + iy), −i(x + iy)] = [T (x + iy), x + iy]
Thus it can be concluded that (T x, y) = (T y, x), or according to the symmetry
of inner product (T x, y) = (x, T y), namely the operator T is symmetric. Quod erat
demonstrandum.
If T is a self-conjugate operator, then
(T x, y) = (x, T
∗ y) = (x, T y) = (T y, x)
It follows that the self-conjugate operator must be the symmetric operator, but the
symmetric operator is not necessarily the self-conjugate operator. thus, by theorem
7.4.4 the following corollary can be obtained:
Corollary 7.4.1 The necessary and sufficient conditions that the operator T is the
self-conjugate operator are: The inner product (T x, x) is a real number.
Theorem 7.4.5 Let T 1 and T 2 be both self-conjugate operators, then the necessary
and sufficient conditions that T 1 T 2 is also a self-conjugate operator are T 1 T 2 = T 2 T 1 .
Proof The necessity. If T 1 T 2 is a self-conjugate operator, then for any x, y ∈ H ,
there is
(T 1 T 2 x, y) = (x, T 1 T 2 y)
Since T 1 and T 2 are both self-conjugate operators, and they are the bounded linear
operator from the Hilbert space H to the Hilbert space H , there is
(T 1 T 2 x, y) = (T 2 x, T 1 y) = (x, T 2 T 1 y)