Question regarding Hermitian property of Operators

Question

I am a beginner in quantum mechanics and some confusion came up in the notations while reading about the Hermitian property of operators. I was told that an operator $\hat {A}$ is Hermitian if the following holds true: $$\int \psi_i ^* \hat {A} \psi_j d\tau = \int (\psi_j^* \hat {A} \psi_i )^*d\tau =\int(\hat {A} \psi_i)^*\psi_jd\tau $$ But I think that this is only valid for operators expressed in usual mathematical notation,not in matrix form.
In matrix form, I think $*$ should be replaced by $\dagger$.
Because if we treat '$*$' as only the complex conjugate,then in matrix form $\psi ^*\psi$ is not a real number this is not even a matrix multiplication(conjugate matrix into the original matrix)which should technically yeild a finite real number,While $\psi^\dagger\psi$ represents a real number(positive real number) .I am sure I am confusing some notations so please clarify if the hermitivity property in matrix notation is : $$\int \psi ^\dagger \hat {A} \psi d\tau = \int (\hat {A} \psi ^* )^\dagger \psi d\tau $$ is correct in matrix notation and also please confirm if $\int(\psi ^\dagger\psi)d\tau$ represents probability density in matrix notation.

Answer 1

The way you define whether something is Hermitian depends on the inner product. What you're giving there is the inner product used for continuous functions in $L^2$. Generally, what you want is that $\langle A\psi|\psi '\rangle = \langle \psi |A \psi '\rangle $. What this condition "looks like" depends on the system (infinite dimensional or not).

If you view $\hat{A}$ as a matrix instead, the condition is exactly what you suggested. For finite-dimensional Hilbert spaces, the bras are naturally defined as column vectors $\langle \psi |= (\psi _1 , \psi _2, \dots )$ and the kets correspondingly as row vectors. So in that case, we want something like $(\langle \psi | \hat{A}^\dagger)|\psi '\rangle = \langle \psi | (A|\psi '\rangle)$, or in other words, that $\hat{A}=\hat{A}^\dagger$.

You can check that this works for discrete versions of differential operators, for example. In that case, we set $|\psi \rangle = (\psi _1 , \psi _2, \dots )$ where $\psi _1$ is the wave function at point $x_1$ and so on. The derivative is then a matrix (using $\frac{df}{dx} = \frac{f(x+h)-f(x-h)}{2h}$ for some finite h)

$$\hat{D} = \frac{1}{2h}\begin{bmatrix} 0 & 1 & 0 & \dots \\ -1 & 0 & 1 & \dots\\ 0 & -1 & 0 & \ddots \\ \vdots &\vdots & \ddots & \ddots \end{bmatrix}$$

Can you see why this is the finite difference derivative matrix? It's clearly not Hermitian. Is it hermitian if you multiply it with i?

In the continuous case, one has to do a partial integration to see if the self-adjointness condition is fulfilled; this is explained in, for example, this answer.

Answer 2

It looks like you are mixing notations. Let us clarify a few things.

Let $\mathcal{H}$ be our Hilbert space. Let $\hat{A}$ be an operator over our Hilbert space, i.e., $\hat{A}: \mathcal{H} \rightarrow \mathcal{H}$. The operator $\hat{A}$ is hermitian by definition when $$\langle \hat{A} \psi, \varphi \rangle = \langle\psi, \hat{A}\varphi\rangle \tag{1}$$ for any $\psi, \varphi \in \mathcal{H}$. Note that $\langle . , . \rangle$ is our usual inner product.

The definition (1) is written as abstractly as possible. We are NOT representing the operator as a matrix nor vectors as columns or rows.

In bra-ket notation, (1) looks like: $$(\hat{A} \lvert \psi \rangle)^{\dagger}\lvert\varphi\rangle = \langle\psi\lvert\hat{A}^{\dagger}\lvert\varphi\rangle = \langle\psi\lvert\hat{A}\lvert\varphi\rangle = (\lvert\psi\rangle)^{\dagger}\;\hat{A}\lvert\varphi\rangle$$ which is more along the lines of how you are thinking.

I think the problem is that the notation you are using in your post is probably referring to states not as vectors, but as functions $\psi(\tau)$. In this case an operator is Hermitian when $$\int \psi^*\hat{A}\varphi \ d\tau = \int (\hat{A} \psi)^*\varphi \ d\tau.$$