# Adjoint of an operator

1. Jun 16, 2005

### Feynmanfan

Could you help me calculate the adjoint of this operator?: Â= x d/dx

I don't know where to start from. I am asked to say if it can be an observable. Well if I find that the adjoint is the same then the operator is Hermitian, right? It can only be observable if it's hermitian.

Thanks!

2. Jun 16, 2005

### matt grime

What's the inner product? without knowign that the question is impossible.

It is the operator L such that <Lf,g> = <f,xg'> by definition, so just do it ie work it out (i'm guessing it's some L^2 kind of space).

3. Jun 16, 2005

### dextercioby

I can assume you're dealing with a rigged Hilbert space in which the nuclear subspace is the the set of tempered distributions of $\mathbb{R}$.

The basis in the Hilbert space is made up of Hermite polynomials and the scalar product is defined

$$(\phi,\psi)=:\int \bar{\phi}(x) \psi (x) \ dx$$.

Here are 3 nice articles reccomended by mr.Bogolubov

Kristensen,P.,Mejlbo L.,Poulsen,E.T. (1964-1967)

"Tempered Distributions in Infinitely Many Dimensions:I Canonical Field Operators",Commun.Math.Phys.,1,175.

"II.Displacement Operators",Math.Scand.,14,129.

"III.Linear Transformations of Field Operators",Commun.Math.Phys.,6,29.

Daniel.

4. Jun 16, 2005

### dextercioby

As for being an observable or not,it must be
1.linear.
2.densly-defined.
3.self-adjoint.

Daniel.

5. Jun 16, 2005

### StatusX

The operator can be written as $\frac{i}{\hbar}\hat x \hat p$, so try taking the adjoint of that (you can ignore h if you want, or set it to 1). Remember x and p are self adjoint, take the complex conjugate of the constant out front, and, just like with transposed matrices, the order of the operators needs to be reversed, ie, (Adj(AB)=Adj(B)Adj(A)).

6. Jun 16, 2005

### dextercioby

That's not true in this case.

$$\hat{p}^{\dagger}\hat{x}^{\dagger}\subset\left(\hat{x}\hat{p}\right)^{\dagger}$$

,because both operators are unbounded on the Hilbert space of states (spanned by Hermite polynomials,check post #3) and therefore can be defined only on an everywhere dense subset of $\mathcal{H}$,called nuclear subspace.

Things are not really easy.

Daniel.

7. Jun 16, 2005

### StatusX

I'm not familiar with nuclear subspaces, but I tried my method and it seemed to work, if you define the adjoint here as the operator such that:

$$\int dx f^{*}(x) \hat L ( g(x) ) = \int dx (\hat L^{\dagger} (f(x)))^{*} g(x)$$

where the integral is taken over the possible range of the particle (usually -$\infty$ to +$\infty$), and the functions are assumed to vanish at the boundaries. I doubt if the question intended for anything more general than this.

Last edited: Jun 16, 2005
8. Jun 16, 2005

### dextercioby

The question asked to compute the adjoint.I think it's pretty clear.One has to apply the definition of the adjoint,to see whether the adjoint exists and under what circumstances.

The problem is not easy,if it were,we'd not be doing mathematics here,but something totally different.

Daniel.

9. Jun 16, 2005

### AKG

How would you do this? It would seem that we need to have:

$$f^{*}(x) \hat L ( g(x) ) = (\hat L^{\dagger} (f(x)))^{*} g(x)$$

$$xf^{*}(x) g'(x) = (\hat L^{\dagger} (f(x)))^{*} g(x)$$

$$\left (\frac{xf^{*}(x) g'(x)}{g(x)}\right )^{*} = \hat L^{\dagger} (f(x))$$

$$\left (\frac{If^{*}g'}{g}\right )^{*} = \hat L^{\dagger} (f)$$

where I is the identity mapping. This doesn't seem to be well-defined, since $L^{\dagger} (f)$ depends on g, so I assume this is not the right way to find $L^{\dagger}$. So how do you do it?

10. Jun 16, 2005

### StatusX

I used my suggestion above (using the x and p operators) to derive the adjoint, and then I verified that it satisfied that condition. In general, the integrands are not equal, and in fact my idea only works if you assume the function vanishes at the endpoints.

Last edited: Jun 16, 2005
11. Jun 17, 2005

### matt grime

i think, daniel, that you're making it far more complicated than it should be given it was homework.

it ought to be quite straight forward, if we had full information, such as what the space is and the inner product.

12. Jun 17, 2005

### dextercioby

Maybe so,but i made my point.

Daniel.

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