Matrix representation of operators

In summary, the conversation discusses the representation of linear operators with matrices and a wave function with a column vector. It is proven that the equation ##\hat{A}(\hat{B}u)=(\hat{A}\hat{B})u## does not hold, revealing a mistake in the calculation of the RHS. The mistake lies in the abuse of notation for ##\hat{B}##, which should be represented by ##\Psi(x)\mapsto x\Psi(x)##. It is also mentioned that all linear operators can be represented by matrices, not just Hermitian ones.
  • #1
Happiness
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Let the operators ##\hat{A}## and ##\hat{B}## be ##-i\hbar\frac{\partial}{\partial x}## and ##x## respectively.

Representing these linear operators by matrices, and a wave function ##\Psi(x)## by a column vector u, by the associativity of matrix multiplication, we have

##\hat{A}(\hat{B}##u##)##=##(\hat{A}\hat{B})##u.

By the definitions of ##\hat{A}## and ##\hat{B}##, we have

LHS ##= -i\hbar\frac{\partial}{\partial x}\big(x\ \Psi(x)\big) = -i\hbar\big(x\frac{\partial\Psi(x)}{\partial x} + \Psi(x)\big)##

RHS ##= -i\hbar\Psi(x)##

LHS ##\neq## RHS, a contradiction. Where is the mistake?

The extract from a textbook below only talks about the matrix representation of a Hermitian operator. But is it true that all linear operators can be represented by matrices, not just those that are Hermitian?

Screen Shot 2016-01-03 at 6.48.28 am.png
 
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  • #2
Happiness said:
Let the operators ##\hat{A}## and ##\hat{B}## be ##-i\hbar\frac{\partial}{\partial x}## and ##x## respectively.

Representing these linear operators by matrices, and a wave function ##\Psi(x)## by a column vector u, by the associativity of matrix multiplication, we have
...
RHS ##= -i\hbar\Psi(x)##

LHS ##\neq## RHS, a contradiction. Where is the mistake?
The mistake is in the RHS calc. It arises from the abuse of notation of writing ##\hat B## as ##x##. THe correct representation of ##\hat B## is ##\Psi(x)\mapsto x\Psi(x)##. That is, it is the operator that, given a function of ##x##, returns a new function that is the original function, multiplied by x. So we get:
##\hat A\hat B=-i\hbar\partial_x \Big(\Psi(x)\mapsto x\Psi(x)\Big)=
\Psi(x)\mapsto -i\hbar \Big(\Psi(x)\partial_x( x)+x\partial_x\Psi(x)\Big)## which will match the LHS when applied to a specific ket.
The extract from a textbook below only talks about the matrix representation of a Hermitian operator. But is it true that all linear operators can be represented by matrices, not just those that are Hermitian?
Yes. The matrix element given in formula 5.156 is just as meaningful for non-Hermitian operators. It will just be (in most cases) a non-Hermitian matrix.
 
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What is a matrix representation of an operator?

A matrix representation of an operator is a way of representing a mathematical operator as a matrix, where each element of the matrix corresponds to a specific operation on a vector space. This allows for easy manipulation and calculation of the operator's effects on vectors.

Why is matrix representation important in mathematics?

Matrix representation is important in mathematics because it allows for the manipulation and calculation of complex operators in a more simplified and visual way. It also allows for the use of matrix algebra, which is a powerful tool in solving various mathematical problems.

How do you determine the matrix representation of an operator?

The matrix representation of an operator can be determined by first choosing a basis for the vector space, then representing the operator as a linear transformation in terms of that basis. The coefficients of the linear transformation matrix will then be the elements of the matrix representation of the operator.

What is the relationship between a matrix representation and an eigenvalue?

A matrix representation of an operator can help determine the eigenvalues of the operator. The eigenvalues are the values that, when multiplied by a vector, produce a scalar multiple of that vector. These eigenvalues can be found by solving the characteristic equation of the matrix representation.

Can the matrix representation of an operator change?

Yes, the matrix representation of an operator can change depending on the chosen basis for the vector space. Different bases can result in different matrix representations of the same operator. However, the eigenvalues of the operator will remain the same regardless of the chosen basis.

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