First order perturbation derivation

[spacetimedude]

In lectures, I learned that in first order perturbation, [itex]\hat{H}_0[/itex] term cancels with [itex]E_0[/itex] term because [itex]\hat{H}_0[/itex] is Hermitian. What property does Hermitian operators hold that cancels with the unperturbed energy?

 

[PeterDonis]

In lectures

Can you give an online reference to the lectures? Or at least information about what lectures (which school, which class, which textbook are you using, etc.)?

 

[spacetimedude]

Can you give an online reference to the lectures? Or at least information about what lectures (which school, which class, which textbook are you using, etc.)?

http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect17.pdf
I am not sure if you can access the page but it is on the third page of the notes.

 

[PeterDonis]

Ok, so the cancellation in question is due to the equality:

$$
\langle \psi_0 \vert \hat{H}_0 \vert \psi_1 \rangle = E_0 \langle \psi_0 \vert \psi_1 \rangle
$$

Since ##\psi_0## is an eigenfunction of ##\hat{H}_0##, we know that ##\hat{H}_0 \vert \psi_0 \rangle = E_0 \vert \psi_0 \rangle##. But the fact that ##\hat{H}_0## is Hermitian let's us flip this around so that ##\hat{H}_0## operates to the left instead of to the right, i.e., we have ##\langle \psi_0 \vert \hat{H}_0 = E_0 \langle \psi_0 \vert##. And that is exactly what we need in order for the equality above, that justifies the cancellation in the first-order equation, to hold.

 

[spacetimedude]

Ok, so the cancellation in question is due to the equality:

$$
\langle \psi_0 \vert \hat{H}_0 \vert \psi_1 \rangle = E_0 \langle \psi_0 \vert \psi_1 \rangle
$$

Since ##\psi_0## is an eigenfunction of ##\hat{H}_0##, we know that ##\hat{H}_0 \vert \psi_0 \rangle = E_0 \vert \psi_0 \rangle##. But the fact that ##\hat{H}_0## is Hermitian let's us flip this around so that ##\hat{H}_0## operates to the left instead of to the right, i.e., we have ##\langle \psi_0 \vert \hat{H}_0 = E_0 \langle \psi_0 \vert##. And that is exactly what we need in order for the equality above, that justifies the cancellation in the first-order equation, to hold.

I see. Is it okay to think that the ##\hat{H}_0## is outside of the bra-ket or is it notationally wrong and operators always have to be kept inside?

 

[PeterDonis]

Is it okay to think that the ##\hat{H}_0## is outside of the bra-ket or is it notationally wrong and operators always have to be kept inside?

The strict way to do it is to always keep the operators inside either the bra or the ket. In other words, the expression ##\langle \psi_0 \vert \hat{H}_0 \vert \psi_1 \rangle## is formally ambiguous; it could mean either ##\langle \psi_0 \vert \hat{H}_0 \psi_1 \rangle##, where the operator is applied to the vector on the right, or ##\langle \hat{H}_0 \psi_0 \vert \psi_1 \rangle##, where the operator is applied to the vector on the left. However, many texts are not this strict and write the operator between the bra and the ket, leaving it to context to show which operation is meant--or relying on the fact that in many cases, such as the one under discussion, it doesn't matter.

 

[spacetimedude]

The strict way to do it is to always keep the operators inside either the bra or the ket. In other words, the expression ##\langle \psi_0 \vert \hat{H}_0 \vert \psi_1 \rangle## is formally ambiguous; it could mean either ##\langle \psi_0 \vert \hat{H}_0 \psi_1 \rangle##, where the operator is applied to the vector on the right, or ##\langle \hat{H}_0 \psi_0 \vert \psi_1 \rangle##, where the operator is applied to the vector on the left. However, many texts are not this strict and write the operator between the bra and the ket, leaving it to context to show which operation is meant--or relying on the fact that in many cases, such as the one under discussion, it doesn't matter.

Thank you very much!