Square of a hermitian operator in matrix form

In summary, if we have a hermitian operator Q and its matrix representation [Q], then we can determine that [Q2] = [Q]2. This is similar to the case of p2 for a harmonic oscillator, where p=ic(a+-a-) and [p2]=[p]2. This relationship holds for any linear operator, not just hermitian ones. However, for non-hermitian operators, the operator may not commute with its Hermitian transpose, so the order of multiplication matters. The result [A°B]=[A][B] holds due to the definition of matrix multiplication. This relationship is further explained in post #3 of the linked thread.
  • #1
Arijun
21
1
If we have a hermitian operator Q and we know it's matrix representation [Q], does that mean that [Q2] = [Q]2?
For example, I'm pretty sure that's the case for p2 for a harmonic oscillator. We have p=ic(a+-a-) and so
p2=c2(a+-a-)(-a++a-)*=c2(a+-a-)(a+-a-)=p p
Which tells us that [p2]=[p]2. But could we do that with any hermitian operator?
 
Physics news on Phys.org
  • #2
Yes. In general, if [itex]C = AB[/itex], then [itex]\left[ C \right] = \left[ AB \right] = \left[ A \right] \left[ B \right][/itex] for any [itex]A[/itex] and [itex]B[/itex] (not just for hermitian operators).
 
  • #3
But if it's not hermitian, then the operator may not commute with its Hermitian transpose (right?), so how would we know whether [C2]=[C] [C]* or [C]* [C], and generally wouldn't there be some commutator relationship in there?
 
  • #4
C2 means CC (the composition of C with C), not C*C or CC*.

The result [A°B]=[A] follows immediately from the definition of matrix multiplication. Actually, I think matrix multiplication is defined the way it is precisely to ensure that this result holds.

$$
\begin{align}
[A\circ B]_{ij} &=((A\circ B)e_j)_i=(A(Be_j))_i =(A(Be_j)_k e_k)_i =(A_{kj} e_k)_i\\
&=(Ae_k)_i _{kj}=[A]_{ij}_{kj} = ([A])_{ij}
\end{align}
$$

If you're having difficulties following this, read the post I linked to in the quote below. Ask if it's still not clear.

Fredrik said:
The relationship between linear operators and matrices is explained e.g. in post #3 in this thread. (Ignore the quote and the stuff below it).
 

1. What is a Hermitian operator in matrix form?

A Hermitian operator is a linear operator on a complex vector space that is equal to its own conjugate transpose. In matrix form, a Hermitian operator has the property that its elements satisfy the condition aij = aji*.

2. What is the square of a Hermitian operator in matrix form?

The square of a Hermitian operator in matrix form is obtained by multiplying the matrix by itself. This operation is also known as squaring the matrix.

3. What is the significance of the square of a Hermitian operator in matrix form?

The square of a Hermitian operator in matrix form is important in quantum mechanics as it represents the observable quantity of a physical system. It also has a number of important properties, such as being a positive semi-definite matrix.

4. How is the square of a Hermitian operator calculated?

The square of a Hermitian operator in matrix form can be calculated by first finding the conjugate transpose of the matrix, then multiplying it by the original matrix. This results in a new matrix that represents the square of the Hermitian operator.

5. What is the difference between a Hermitian operator and a non-Hermitian operator?

A Hermitian operator is equal to its own conjugate transpose, while a non-Hermitian operator is not. This means that a Hermitian operator has real eigenvalues and its eigenvectors are orthogonal, while a non-Hermitian operator may have complex eigenvalues and its eigenvectors are not orthogonal.

Similar threads

Replies
2
Views
699
Replies
16
Views
2K
Replies
10
Views
1K
Replies
5
Views
1K
  • Quantum Physics
Replies
7
Views
1K
  • Quantum Physics
Replies
14
Views
2K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
67
Views
4K
  • Quantum Physics
Replies
2
Views
14K
Back
Top