# Quantum Mechanics : Operators

1. Oct 6, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
Consider the following operators
a) Reflection: $$\hat{I}\Psi(x)=\Psi(-x)$$ , $$x\in(-\infty,\infty)$$
b) Translation: $$\hat{T_{a}}\Psi(x)=\Psi(x+a)$$, $$x\in(-\infty,\infty)$$
c) $$\hat{M_{c}}: \hat{M_{c}}\Psi(x) = \sqrt{c}\Psi(x)$$
d) $$\hat{c}\Psi(x)= (\Psi(x))^*$$
e) $$i\frac{\partial}{\partial x}, x\in(-\infty,\infty)$$
f) $$i\frac{\partial}{\partial r}, r\in[0,\infty)$$ where r is the radial variable

Are the operators linear? Find operators which are complex conjugate, transposed, Hermitian conjugate, and inverse with respect to the given operators

2. The attempt at a solution

Ok so the book (Pauli's) says If F is a linear operator than corresponding to every eigenfunction $u_{n}$ there is an expansion

$$(Fu_{n}) ~\sum_{k} u_{k} F_{kn}$$
with $$F_{kn} = \int u_{k}^* ((Fu_{n}) dq$$

For the reflection operator then would writing something like this be sufficient?
$$\hat{I}\Psi(x)=\sum_{n}u_{n} \Psi_{n}(x) I_{kn} = \Psi(-x)$$

i alwyas thought that to prove the linearity of an operator we would have to do something like this

For two function $\Psi_{1}$ and $\Psi_{2}$
$$\Hat{I} (\Psi_{1} + \Psi_{2})(x) = \Psi_{1}(-x) +\Psi_{2}(-x)$$ which means it is linear
is there something else i nthe first method that i have not pointed out??

i am actually not sure how to find the inverse of an operator... it doesnt seem to be given in the textbook for my class (Griffiths)

This is for an assignment but its pointless doing it for the assignment... i need to find the explicit formula (i think that one has to do the an integral with integrating by parts and so on) Just point out the book in which i need to look thats all i ask

thanks

2. Oct 7, 2007

### stunner5000pt

is the Hermitian conjugate computed in the following way

$$\int \Psi^* Q^{\dagger}\Psi dx = \int (Q\Psi)^* \Psi dx$$

and the complex conjugate computed like this

$$\int \Phi^* A \Psi dx = \int (A\Phi)^* \Psi x$$

but how to compute the coplex conjugate ... do i simply just change the sing in front of the i in the operator??

what about the transposed? If it was a matrixn no problemn but a one dimensional operator... ?

3. Oct 7, 2007

### dextercioby

Can you prove that all the operators, when defined on appropriate subsets of a Hilbert space (called domains), are linear ?

4. Oct 7, 2007

### stunner5000pt

ok to prove some operator is linear am i not supposed to go about it this way...

for the reflection operator

$$\Hat{I} (\Psi_{1} + \Psi_{2})(x) =(\Psi_{1} + \Psi_{2})(-x) = \Psi_{1}(-x) + \Psi_{2}(-x)$$
which means it is linear... correct??

5. Oct 8, 2007

### stunner5000pt

i believe i made a mistake here

there should be scalars with the Psis that is
$$\Hat{I} (a_{1}\Psi_{1} +a_{2} \Psi_{2})(x) =(a_{1}\Psi_{1} + a_{2}\Psi_{2})(-x) = a_{1}\Psi_{1}(-x) + a_{2}\Psi_{2}(-x)$$

this is fine but how to find the complex conjugate and transposed operators??

to find the complex conjugate of the reflection one... is it simply itself since there is no "i" involved??

6. Oct 8, 2007

### stunner5000pt

i found the definitions which our prof had given us they looked like this

$$\int \Psi^* \hat{Q} \Psi dx = \int \hat{Q}^* \Psi^* \Psi dx$$

$$\int \Psi^* \hat{Q} \Psi dx = \int \hat{Q}^* \Psi^* \Psi dx$$ complex conjugate ?
$$\int \Psi^* \hat{Q} \Psi dx = \int \Psi^* (\hat{Q}^*)^T \Psi dx$$ tranpose

and
$$\int \Psi^* \hat{Q} \Psi dx = \int \hat{Q}^{dagger} \Psi^* \Psi dx$$ for hermitian conjugate

but i cannot find the inverse

are these defintions correct?? the source was from someone's notes and as such they may be flawed

7. Oct 9, 2007

### stunner5000pt

are the definitons right?? They look circular to me...

but ok lets try to find the complex conjugate for the reflection operator

$$\int \Psi^* \hat{Q} \Psi dx = \int \hat{Q}^* \Psi^* \Psi dx$$
$$\int \Psi^* \hat{I} \Psi dx = \int \Psi(x)^* \Psi(-x) dx$$

stuck now... would i have to use integration by parts to solve this?? What can be said about the reflection of a wavefunction??