# Finding adjoint of an operator

In summary: Yes, that is correct. You have shown that the adjoint of the operator A is also an integral operator, with the function g replacing the function f. In summary, the adjoint of the operator A:L^2[0,1] -> L^2[0,1] defined by (Af)(x) = integ from (0 to x) f(t)dt is another integral operator with the function g replacing the function f, as shown by the formula <Af, g>= <f, A*g>. This is the unique operator that satisfies this property.
Find the adjoint of the operator A:L^2[0,1] -> L^2[0,1] defined by (Af)(x) = integ from (0 to x) f(t)dt

so from my notes it says: the operator A* is called the adjoint of A if:
<Ax, y> = <x, A*y> for all x, vE H

i am not sure how to do this, and need to know how to do it for a test

Think about what the inner product in this case is, and keep in mind that the operator A* that satisfies <Ax,y>=<x,A*y> for all x and y is unique.

so inner product is <f,g>=integral f(x)g(x) dx

Okay, so <Af, g>= <f, A*g> is $\int (Af)g dx= \int f(Ag)dx$

Since $Af= \int_0^x f(t)dt$ that becomes
$$\int\left[\int_0^x f(t)dt\right]g(x)dx= \int f(x)\left[A*g(x)\right]dx$$

You might try doing the left side by "integration by parts" letting $u= \int_0^x f(t)dt$ and dv= g(x)dx

i am trying to do the lhs as you said
but i am getting stuck, i have not done integration by parts like this before.
so for u = integ (0 to x) f(t) dt
u' = f(t) ? but what happens to the limits of integration, do they just carry along.
and for dv = g(x) dx,
dv/dx = g(x)
v = integ (g(x) dx)
is that correct?
but what has changed, since there is still a double integral on LHS, and what should i do with the RHS?

i am trying to do the lhs as you said
but i am getting stuck, i have not done integration by parts like this before.
so for u = integ (0 to x) f(t) dt, u' = f(t) ?
No, du= f(x)dx. The t is a dummy variable.

but what happens to the limits of integration, do they just carry along.
Remember the formula for integration by parts?
$$\int_a^b udv= uv|_a^b- \int_a^b v du$$

and for dv = g(x) dx,
dv/dx = g(x)
v = integ (g(x) dx)
is that correct?
It might be better to write
$$v= \int_0^x g(t)dt$$

but what has changed, since there is still a double integral on LHS, and what should i do with the RHS?
What has changed is the the "inner integral" on the left is now an integral of g rather than f. Compare the right and left sides and you should conclude that A*g must also be an integral.

ok, after the udv part i got
integ (0-x) f(t)dt integ (0-x) g(t)dt - integ (integ (0-x) g(t)dt ) f(x) dt
I see what you are talking about the inner integral, but can i just conclude after this step that A*g must also be an integral?

## 1. What is an adjoint operator?

An adjoint operator is a mathematical concept used in linear algebra to represent the relationship between two vector spaces. It is the dual of a linear operator, and it allows for the transformation of vectors from one space to another.

## 2. How is the adjoint of an operator calculated?

The adjoint of an operator is calculated by taking the conjugate transpose of the operator's matrix representation. This involves taking the complex conjugate of each element in the matrix and then transposing the matrix.

## 3. What is the importance of finding the adjoint of an operator?

Finding the adjoint of an operator is important because it allows us to perform calculations involving inner products and orthogonality in vector spaces. It also allows us to solve problems in quantum mechanics and other fields that involve linear transformations.

## 4. How is the adjoint of an operator used in quantum mechanics?

In quantum mechanics, the adjoint of an operator is used to calculate the expectation value of an observable in a quantum state. It is also used in the mathematical formulation of the Schrödinger equation, which describes the time evolution of a quantum system.

## 5. Can every operator have an adjoint?

No, not every operator has an adjoint. An operator must be bounded and defined on a Hilbert space in order for it to have an adjoint. Additionally, the adjoint may not exist if the operator is not self-adjoint or if the underlying vector space is not complete.

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