Undergrad Hermitian Operators: Referencing Griffiths

[WWCY]

I have a few issues with understanding a section of Griffiths QM regarding Hermitian Operators and would greatly appreciate some help.

It was first stated that,

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

and because expectation values are real,

$\langle Q \rangle = \langle Q \rangle ^*$

By invoking $\langle f|g \rangle = \langle g|f \rangle ^*$ (1)

$\langle \Psi | \hat{Q} \Psi \rangle = \langle \Psi | \hat{Q} \Psi \rangle ^* = \langle \hat{Q} \Psi| \Psi \rangle$.

Which I so far, seem to understand. But then it was stated that,

$\langle f| \hat{Q}g \rangle = \langle \hat{Q}f|g \rangle$ was the condition for a Hermitian operator.

Up till now my understanding (which seems plainly wrong) was as follows,

$\langle f | \hat{Q} g\rangle = \langle f | \hat{Q} g\rangle ^* = \langle \hat{Q}g|f \rangle$

where step 2 to 3 involved the flipping of functions as seen in (1).

Could anyone explain how it's supposed to work? Assistance is greatly appreciated!

P.S. It would be nice if explanations could be kept simple, I have not worked up till all things "eigen" (next section of book) as of yet. Thanks!

 

[kuruman]

This is always true $\langle f|g \rangle = \langle g|f \rangle ^*$.
Now consider $\langle \Phi | \hat{Q} \Psi \rangle$. Define $| \hat{Q} \Psi \rangle = | \Omega \rangle$. Then
$\langle \Phi |\Omega\rangle = \langle \Omega| \Phi \rangle ^*$ is always true which means that $\langle \Phi |\hat{Q}\Psi\rangle = \langle \hat{Q} \Psi| \Phi \rangle ^*$ is also always true.

However, if it is also true that $\langle \Phi | \hat{Q}\Psi\rangle = \langle \hat{Q} \Psi| \Phi \rangle$, then $\hat{Q}$ is hermitian. The converse is also true. If $\hat{Q}$ is hermitian, then $\langle \Phi | \hat{Q}\Psi\rangle = \langle \hat{Q} \Psi| \Phi \rangle$.

That's how it works.

 

[WWCY]

This is always true $\langle f|g \rangle = \langle g|f \rangle ^*$.
Now consider $\langle \Phi | \hat{Q} \Psi \rangle$. Define $| \hat{Q} \Psi \rangle = | \Omega \rangle$. Then
$\langle \Phi |\Omega\rangle = \langle \Omega| \Phi \rangle ^*$ is always true which means that $\langle \Phi |\hat{Q}\Psi\rangle = \langle \hat{Q} \Psi| \Phi \rangle ^*$ is also always true.

However, if it is also true that $\langle \Phi | \hat{Q}\Psi\rangle = \langle \hat{Q} \Psi| \Phi \rangle$, then $\hat{Q}$ is hermitian. The converse is also true. If $\hat{Q}$ is hermitian, then $\langle \Phi | \hat{Q}\Psi\rangle = \langle \hat{Q} \Psi| \Phi \rangle$.

That's how it works.

Thanks @kuruman , that helped clear things up a bit. However I still don't get the flow of logic behind this$\langle f| \hat{Q}g \rangle = \langle \hat{Q}f|g \rangle$. What allows us to swap only the operator?

 

[DrClaude]

I don't like that "operator in ket" notation, but let's simply take it as a shorthand for $|\hat{Q}g \rangle = \hat{Q} |g \rangle$. Then, for $\hat{Q} = \hat{Q}^\dagger$
$$
\begin{align*}
\langle f | \hat{Q}g \rangle &= \langle f | \hat{Q} |g \rangle \\
&= \left[\langle f | \hat{Q} |g \rangle^\dagger \right]^\dagger \\
&= \left[\langle g | \hat{Q}^\dagger |f \rangle \right]^\dagger \\
&= \left[\langle g | \hat{Q} |f \rangle \right]^\dagger \\
&= \left[\langle g | \hat{Q} f \rangle \right]^\dagger \\
&= \langle \hat{Q} f | g \rangle
\end{align*}
$$

 

[WWCY]

I don't like that "operator in ket" notation, but let's simply take it as a shorthand for $|\hat{Q}g \rangle = \hat{Q} |g \rangle$. Then, for $\hat{Q} = \hat{Q}^\dagger$
$$
\begin{align*}
\langle f | \hat{Q}g \rangle &= \langle f | \hat{Q} |g \rangle \\
&= \left[\langle f | \hat{Q} |g \rangle^\dagger \right]^\dagger \\
&= \left[\langle g | \hat{Q}^\dagger |f \rangle \right]^\dagger \\
&= \left[\langle g | \hat{Q} |f \rangle \right]^\dagger \\
&= \left[\langle g | \hat{Q} f \rangle \right]^\dagger \\
&= \langle \hat{Q} f | g \rangle
\end{align*}
$$

So based on post 2, $\langle \hat{Q} f | g \rangle = langle \hat{Q} g | f \rangle$?

Also, why does $\hat{Q} = \hat{Q}^\dagger$? If we were talking about the momentum operator, does it not change its sign due to the $i$ it has?

Thank you for your assistance and patience.

 

[DrClaude]

So based on post 2, $\langle \hat{Q} f | g \rangle = \langle \hat{Q} g | f \rangle$?

I would say no, it's $\langle \hat{Q} f | g \rangle = \langle \hat{Q} g | f \rangle^\dagger$. What @kuruman wrote is not obvious to me; maybe he can clarify.

Also, why does $\hat{Q} = \hat{Q}^\dagger$?

Because we are discussing Hermitian operators:

I have a few issues with understanding a section of Griffiths QM regarding Hermitian Operators

If we were talking about the momentum operator, does it not change its sign due to the $i$ it has?

See https://www.physicsforums.com/threads/hermitian-momentum-operator.742281/

 

[kuruman]

I too dislike "operators in kets" notation, but I stuck with it to reply consistently with OP's initial question. I tried to sort this out thinking of the brakets as integrals. In terms of integrals, an operator $\hat{Q}$ is hermitian iff
$$\int{\Phi^*(\hat{Q}\Psi)~dx}=\int{(\hat{Q}\Phi)^*\Psi~dx}$$
which in post #2 I incorrectly translated to "operators in kets" notation. I should have written
$$\langle \Phi | \hat{Q}\Psi\rangle = \langle \hat{Q} \Phi| \Psi \rangle$$
Sorry about the confusion I may have caused.

 

[WWCY]

Thanks for the clarification!

However @DrClaude , I couldn't follow the arguments in the thread as it was still too technical for me.

Can I take it that $\hat{Q} = \hat{Q}^*$ is a condition for a hermitian operator?

 

[DrClaude]

Can I take it that $\hat{Q} = \hat{Q}^*$ is a condition for a hermitian operator?

Generally speaking, a Hermitian operator follows $\hat{Q} = \hat{Q}^\dagger$. What the other thread pointed out is that you have to be careful what the $\dagger$ (Hermitian conjugation) does, as in some cases it is not simply equal to complex conjugation (as for the momentum operator).

 

[kuruman]

Let me attempt to clarify with a couple of concrete examples. Suppose you were asked to show that an operator is Hermitian. What do you need to do? The answer is "it depends on the operator." If the operator can be represented by a matrix, such as spin, then you need to show that the matrix representing the operator is self-adjoint. OK what does that mean? It means you need to show that $\hat{Q} = \hat{Q}^\dagger$. OK and what does that mean. It means that if you interchange rows and columns of the matrix and at the same time you take the complex conjugate of the elements, you end up with the same matrix.
Example 1. Consider two matrices
$$M_1=\begin{pmatrix}
0 &i \\
i & 0
\end{pmatrix}
~~~M_2=\begin{pmatrix}
0 &-i \\
i & 0
\end{pmatrix}
$$Note that
$M_1^{\dagger}=\begin{pmatrix} 0 &-i \\ -i & 0 \end{pmatrix}$, from which $M_1\neq M_1^{\dagger}$, therefore $M_1$ is not Hermitian. If you apply the same test to $M_2$, you will see that it is indeed Hermitian.
Example 2. To show that an operator such as momentum is Hermitian, you cannot use the matrix representation. You need to show that
$$\int_{- \infty}^{\infty}{\Phi^*(x) \left( \frac{\hbar}{i} \frac{\partial \Psi(x)}{\partial x} \right) ~dx}=\int_{- \infty}^{\infty}{\left( \frac{\hbar}{i} \frac{\partial \Phi(x)}{\partial x} \right)^*\Psi(x)~dx}$$
This can be done using integration by parts (twice) and assuming that $\Phi(x)$ and $\Psi(x)$ are normalizable and don't do anything horrible as $x \rightarrow \pm \infty$. Operator $\hat{x}$ is clearly Hermitian because it's real.

There is more that can be said about all this, but I will stop here. It should become clearer to you as you delve deeper into QM and develop some familiarity.

On Edit: When I taught QM out of Griffiths, I introduced Hermitian operators and their properties using integrals and before introducing Dirac notation. This seemed to be less confusing to students.

 

[WWCY]

Generally speaking, a Hermitian operator follows $\hat{Q} = \hat{Q}^\dagger$. What the other thread pointed out is that you have to be careful what the $\dagger$ (Hermitian conjugation) does, as in some cases it is not simply equal to complex conjugation (as for the momentum operator).

Example 2. To show that an operator such as momentum is Hermitian, you cannot use the matrix representation. You need to show that
$$\int_{- \infty}^{\infty}{\Phi^*(x) \left( \frac{\hbar}{i} \frac{\partial \Psi(x)}{\partial x} \right) ~dx}=\int_{- \infty}^{\infty}{\left( \frac{\hbar}{i} \frac{\partial \Phi(x)}{\partial x} \right)^*\Psi(x)~dx}$$
This can be done using integration by parts (twice) and assuming that $\Phi(x)$ and $\Psi(x)$ are normalizable and don't do anything horrible as $x \rightarrow \pm \infty$. Operator $\hat{x}$ is clearly Hermitian because it's real.

Thank you both for your responses.

Would I then be right in saying that I should test operators by working them out in integral form (for now)?

 

[kuruman]

Would I then be right in saying that I should test operators by working them out in integral form (for now)?

For now yes, until you go deep in chapter 3 of Griffiths where he starts using Dirac notation.

 

[vanhees71]

Again a warning: It's not sufficient for an operator representing an observable to be Hermitean. It must be even self-adjoint. For details, see the Marvelous pedagogical papers

F. Gieres, Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys., 63 (2000), p. 1893.
http://arxiv.org/abs/quant-ph/9907069

G. Bonneau and J. Faraut, Self-adjoint extensions of operators and the teaching of quantum mechanics, Am. Jour. Phys., 69 (2001), p. 322.
http://arxiv.org/abs/quant-ph/0103153

It seems to me, Griffiths's QT book is sometimes a bit (too?) imprecise, but I've not yet have had a close look at it. So maybe that's an unjustified impression from many discussions related to this textbook in the forums.

 

[WWCY]

Again a warning: It's not sufficient for an operator representing an observable to be Hermitean. It must be even self-adjoint. For details, see the Marvelous pedagogical papers

F. Gieres, Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys., 63 (2000), p. 1893.
http://arxiv.org/abs/quant-ph/9907069

G. Bonneau and J. Faraut, Self-adjoint extensions of operators and the teaching of quantum mechanics, Am. Jour. Phys., 69 (2001), p. 322.
http://arxiv.org/abs/quant-ph/0103153

It seems to me, Griffiths's QT book is sometimes a bit (too?) imprecise, but I've not yet have had a close look at it. So maybe that's an unjustified impression from many discussions related to this textbook in the forums.

Thank you, I'll keep these in the back of my head till I'm able to go through them.

Griffith's book does seem that way to me, but I don't have an alternative that is targetted at novices like myself. Do you have any recommendations?

 

[vanhees71]

My favorite for introductory non-relativistic QM is J. J. Sakurai, Modern Quantum Mechanics (2nd edition).