Is an operator (integral) Hermitian?

[happyparticle]

Homework Statement

Is the operator $\int_{-\infty}^{\infty} |x><x| dx$ Hermitian

Relevant Equations

$\hat{Q} = \hat{Q}^{\dagger}$
$<f|\hat{Q}|g> = <\hat{Q}f|g>$

Knowing that to be Hermitian an operator $\hat{Q} = \hat{Q}^{\dagger}$.
Thus, I'm trying to prove that $<f|\hat{Q}|g> = <\hat{Q}f|g>$.
However, I don't really know what to do with this expression.
$<f|\hat{Q}g> = \int_{-\infty}^{\infty} [f(x)^* \int_{-\infty}^{\infty} |x> <x| dx f(x)] dx$

Am I on the right track?

Thank you

 

[PeroK]

In general $$\bra f Q \ket g = \bra g Q^{\dagger} \ket f^*$$

 

[happyparticle]

I'm facing the same issue. I'm not sure how to deal with the integral as operator.
I'm using Griffith's equation 3.20 $<f|\hat{Q}g> = <\hat{Q}f|g>$

 

[DrClaude]

However, I don't really know what to do with this expression.
$<f|\hat{Q}g> = \int_{-\infty}^{\infty} [f(x)^* \int_{-\infty}^{\infty} |x> <x| dx f(x)] dx$

I can only make sense of the operator $\hat{Q}$ if it operates on kets. So I would keep using the Dirac notation (no $f(x)$ or $g(x)$, but $\bra{f}$ and $\ket{g}$), while introducing the definition of the operator $\hat{Q}$.

 

[hilbert2]

Write an arbitrary state $\left|\psi\right.\rangle$ in the basis of position eigenstates $\left|x\right.\rangle$ and show that the operator in the exercise acts on it like the identity operator, $\hat{Q}\left|\psi\right.\rangle = \left|\psi\right.\rangle$. The identity operator is clearly hermitian, as any state is an eigenstate with eigenvalue 1. The wave function $\psi(x)$ is the multiplier of $\left|x\right.\rangle$ in the integral representation of the arbitrary state in basis of position eigenstates.

 

[happyparticle]

Alright, I still have few questions.

I'm doing what helbert2 said.
$\int_{-\infty}^{\infty} dx |x><x| \psi$

Is $\int_{-\infty}^{\infty} dx (<x|)^t |x> \psi$ = $\int_{-\infty}^{\infty} dx <x|x> \psi = \psi$
correct? If so I'm not sure to understand why $(<x|)^t = |x>$. I just saw that, but without further explanation.

 

[PeroK]

Alright, I still have few questions.

I'm doing what helbert2 said.
$\int_{-\infty}^{\infty} dx |x><x| \psi$

Is $\int_{-\infty}^{\infty} dx (<x|)^t |x> \psi$ = $\int_{-\infty}^{\infty} dx <x|x> \psi = \psi$
correct? If so I'm not sure to understand why $(<x|)^t = |x>$. I just saw that, but without further explanation.

By definition $\bra x$ is the bra associated with the ket $\ket x$.

 

[happyparticle]

Do you mean $|x \rangle \langle x| = \langle x |x \rangle$ ? I'm not so sure to see the link with my questions.

 

[PeroK]

Do you mean $|x \rangle \langle x| = \langle x |x \rangle$ ? I'm not so sure to see the link with my questions.

I definitely don't mean that. The LHS is an operator and the RHS is a number.

You need to revise what all these symbols mean. You start with a ket space, then introduce the idea of a dual space or bra space in Dirac notation.

 

[happyparticle]

I'm not sure to understand what $|x \rangle \langle x|$ is exactly.
for a vector $|x \rangle$

Is it correct to say that
$|x \rangle \langle x|$ is a n x 2 matrix and $\langle x |x \rangle$ is a dot product?

 

[hilbert2]

Alright, I still have few questions.

I'm doing what helbert2 said.
$\int_{-\infty}^{\infty} dx |x><x| \psi$

Is $\int_{-\infty}^{\infty} dx (<x|)^t |x> \psi$ = $\int_{-\infty}^{\infty} dx <x|x> \psi = \psi$
correct? If so I'm not sure to understand why $(<x|)^t = |x>$. I just saw that, but without further explanation.

The wave function $\psi (x)$ of state $\left|\psi\right.\rangle$ is defined as the inner product with a position eigenstate $\left|x\right.\rangle$: $\psi (x) = \langle \left.x\right|\psi\rangle$. This is visible in the calculation

$\displaystyle\langle \left.x\right|\psi\rangle = \langle \left.x\right|\int_{-\infty}^{\infty}\psi (x')\left|x'\right.\rangle dx' = \int_{-\infty}^{\infty}\psi (x')\langle \left.x\right|x'\rangle dx' = \int_{-\infty}^{\infty}\psi (x')\delta (x-x') dx' = \psi (x)$.

A position eigenstate $\langle \left.x\right|$ or $\left|x\right.\rangle$ can be moved in or out of an integral that doesn't integrate over variable $x$. If you apply this and the property $\langle x\left|x'\right.\rangle = \delta (x-x')$ on the quantity $\hat{O}\left|\psi\right.\rangle$, you see that it's the same as $\left|\psi\right.\rangle$.

 

[vanhees71]

Homework Statement:: Is the operator $\int_{-\infty}^{\infty} |x><x| dx$ Hermitian
Relevant Equations:: $\hat{Q} = \hat{Q}^{\dagger}$
$<f|\hat{Q}|g> = <\hat{Q}f|g>$

Knowing that to be Hermitian an operator $\hat{Q} = \hat{Q}^{\dagger}$.
Thus, I'm trying to prove that $<f|\hat{Q}|g> = <\hat{Q}f|g>$.
However, I don't really know what to do with this expression.
$<f|\hat{Q}g> = \int_{-\infty}^{\infty} [f(x)^* \int_{-\infty}^{\infty} |x> <x| dx f(x)] dx$

Am I on the right track?

Thank you

I don't know, how you get two integrals here since you have
$$\langle f|\hat{Q} g \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle f |x \rangle \langle x |g \rangle = \int_{\mathbb{R}} f^*(x) g(x)=\langle f|g \rangle.$$
Since this holds for any $f$ and $g$, this implies that $\hat{Q}=...$?

 

[PeroK]

I'm not sure to understand what $|x \rangle \langle x|$ is exactly.
for a vector $|x \rangle$

Is it correct to say that
$|x \rangle \langle x|$ is a n x 2 matrix and $\langle x |x \rangle$ is a dot product?

You really do need to go back to basics and digest this material.

The outer product $\ket x \bra y$ is an operator that maps a ket $\ket z$ to the ket $(\ket x)\braket {y|z}$.

Note that it's usual to drop the round brackets here, which I've only included for emphasis.

 

[happyparticle]

I don't know, how you get two integrals here since you have
$$\langle f|\hat{Q} g \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle f |x \rangle \langle x |g \rangle = \int_{\mathbb{R}} f^*(x) g(x)=\langle f|g \rangle.$$
Since this holds for any $f$ and $g$, this implies that $\hat{Q}=...$?

I thought because the operator was an integral and the inner product of 2 functions is an integral as well.

I didn't know $\psi (x) = \langle \left.x\right|\psi\rangle$
If I understand griffith (p.90 third edition) this is like a dot product between 2 vectors?

 

[vela]

I thought because the operator was an integral and the inner product of 2 functions is an integral as well.

I didn't know $\psi (x) = \langle \left.x\right|\psi\rangle$
If I understand griffith (p.90 third edition) this is like a dot product between 2 vectors?

You want to recognize that $f(x)$ and $\lvert f \rangle$, while related, are not the same thing and therefore are not interchangeable. $\lvert f \rangle$ denotes a vector while $f(x)$ represents the vector in the position basis. In your initial attempt, you mixed these two types of objects up, which led to the confusing mess you started with.

Say you have two vectors $\mathbf{a}$ and $\mathbf{b}$ in an $n$-dimensional vector space $V$, and you want to calculate $\mathbf{a}\cdot\mathbf{b}$. One way to do that it to first choose an orthonormal basis $\{\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n\}$ and then represent $\mathbf{a}$ and $\mathbf{b}$ by the $n$-tuples $(a_1, a_2, \dots, a_n)$ and $(b_1, b_2, \dots, b_n)$. In other words, you have
\begin{align*}
\mathbf{a} = \sum_{i=1}^n a_i \mathbf{e}_i \\
\mathbf{b} = \sum_{i=1}^n b_i \mathbf{e}_i
\end{align*} Then the dot product would be given by
$$\mathbf{a}\cdot\mathbf{b} = \sum_{i=1}^n a_i b_i.$$

Compare that last expression to the one you learned for $\langle f \vert g \rangle$.
$$\langle f \vert g \rangle = \int f^*(x)g(x)\,dx.$$ $\langle f \rvert$ and $\lvert g \rangle$ corresponds to $\mathbf{a}$ and $\mathbf{b}$. The integration corresponds to summing over $i$. And $f(x)$ and $g(x)$ correspond to the $a_i$'s and $b_i$'s.

Just as $(a_1, a_2, \dots, a_n)$ is merely a representation of $\mathbf{a}$ with respect to the orthonormal basis, $f(x)$ is merely a representation of $\lvert f \rangle$ in the position basis. In other words,
$$\lvert f \rangle = \int \psi(x) \lvert x \rangle\,dx.$$ Compare that expression to the sum above where $\mathbf{a}$ was expanded in terms of the chosen basis. The integral corresponds to the summation; $\psi(x)$ to $a_i$, and $\lvert x \rangle$ to $\mathbf{e}_i$.

Finally, note that
$$\mathbf{e}_i \cdot \mathbf{a} = \mathbf{e}_i \cdot \sum_{j=1}^n a_j \mathbf{e}_j = \sum_{j=1}^n a_j (\mathbf{e}_i \cdot \mathbf{e}_j) = \sum_{j=1}^n a_j \delta_{ij} = a_i.$$ The analogous calculation In Dirac notation is
$$\langle x \vert f \rangle = \langle x \rvert \int f(x')\lvert x' \rangle\,dx' =
\int f(x')\langle x \vert x' \rangle\,dx' =
\int f(x')\delta(x-x')\,dx' = f(x)$$

I glossed over some technicalities since you're generally dealing with complex numbers in quantum mechanics, but the gist should be clear. Dirac notation might look a little strange at first, but once you get used to it, you'll find it's a much more convenient way of expressing the ideas you learned in linear algebra.

 

[vanhees71]

I thought because the operator was an integral and the inner product of 2 functions is an integral as well.

I didn't know $\psi (x) = \langle \left.x\right|\psi\rangle$
If I understand griffith (p.90 third edition) this is like a dot product between 2 vectors?

You should read about the foundations of the theory in terms of the Dirac formulation again. Griffiths's quantum textbook seems to lead to a lot of confusion, as this thread demonstrates once more. I recommend J. J. Sakurai, Modern Quantum Mechanics, Revised Edition

 

[happyparticle]

Alright, thank you for the recommendation.
Thank you all for the help. Sorry if I'm slow to understand, thank you for your patience.

 

[vanhees71]

Don't worry. To understand quantum theory it takes some time!