1. Aug 30, 2006

### enalynned

Im trying to solve problem 2.26 from Griffiths (1st. ed, Intro to Q.M.). Its about the allowed enegy to double dirac potential. I came up with a final equation that is trancedental. (After I separate the even and odd solution of psi.) Am I on the right track?

Please refer to Griffiths book equation number 3.83. Now consider my arguments.

Let lc> = Tlb>, where T is an operator, then <cl = <bl T+, where T+ is the hermitian conjugate of T. One of the property of inner product is.

<alc> = <cla>*
thus
<alTlb> = <blT+la>*

In eqn. 3.83 of Griffiths there is no conjugation when T+ operates on la>...
Does this mean

<blT+la>* = <alT+lb> ?

where * means conjugate
thanks!!!!

Last edited: Aug 31, 2006
2. Aug 31, 2006

### Dr Transport

I would suspect that T is Hermitian...

3. Sep 7, 2006

### enalynned

Hermitian operator is just a special case of adjoints....
Sorry for the late reply ^^

4. Sep 7, 2006

### enalynned

:uhh: Is it always true that
<blT+la>* = <alT+lb>
regardless of T being Hermitian?

5. Sep 7, 2006

### dextercioby

Of course not. Let's say you have

$$\langle b, T^{\dagger}a\rangle$$

That's equal to

$$\langle (T^{\dagger})^{\dagger}b, a\rangle$$

So you'd have to require that the adjoint of the adjoint should exist and moreover

$$T^{\dagger}b=(T^{\dagger})^{\dagger}b \ , \ \forall b\in D(T^{\dagger}) \and b\in D((T^{\dagger})^{\dagger})$$

If that happens, then you can employ Dirac's notation with bars. It's always true that an operator is included in its adjoint's adjoint, but for the adjoint it always have to be checked.

Daniel.

Last edited: Sep 7, 2006
6. Sep 8, 2006

### enalynned

In the first edition of "Introduction to Quantum Mechanics" by Griffiths equation 3.83 he (Griffiths) states that a Hermitian Conjugate is an operator with the property

<alTb>=<T+alb> ... (1)

That is a Hermitian Conjugate (not necessarily Hermitian operator) is a "transformation T+ which, when applied to the first member of an inner product, gives the same result as if T itself had been applied to the second vector."

But from my previous arguments, I obtained

<alTlb>=<blT+la>* ... (2)

if i equate (1) and (2) i will have (if <alTlb> is the same as <alTb>)

<blT+a>*=<T+alb>....

notice that T+ operates on la> now.