I Relation between operators and their variables' signs

[hokhani]

TL;DR Summary

Do differential operators change corresponding to their variables' signs?

Consider the operator $\hat{A} (x,t) =\frac{d}{dx} +f(x,t)$ where $\hat{A}$ is an operator and $f(x,t)$ a function. I don't know which of the following expression is correct for $\hat{A} (-x,-t)$?
first one: $$\hat{A} (-x,-t) =\frac{d}{dx} +f(-x,-t)$$ and the second: $$\hat{A} (-x,-t) =-\frac{d}{dx} +f(-x,-t)$$

 

[otennert]

My take:
$$\hat{A}(-x,-t) = \left.\frac{d}{dx}\right|_{-x} + f(-x,-t)$$

Unless the function acted upon by $\hat{A}(x,t)$ possesses any symmetries, I'd say neither of your suggestions are correct.

 

[PeroK]

TL;DR Summary: Do differential operators change corresponding to their variables' signs?

Consider the operator $\hat{A} (x,t) =\frac{d}{dx} +f(x,t)$ where $\hat{A}$ is an operator and $f(x,t)$ a function. I don't know which of the following expression is correct for $\hat{A} (-x,-t)$?
first one: $$\hat{A} (-x,-t) =\frac{d}{dx} +f(-x,-t)$$ and the second: $$\hat{A} (-x,-t) =-\frac{d}{dx} +f(-x,-t)$$

An operator is a mapping from functions to functions. It's not itself a function of $x$ and $t$.

 

[renormalize]

An operator is a mapping from functions to functions. It's not itself a function of $x$ and $t$.

Of course in general an operator can be a function of the coordinates; e.g., the 3D Laplacian in spherical coordinates:$$\nabla^{2}=\dfrac{1}{r^{2}}\dfrac{\partial}{\partial r}\left(r^{2}\dfrac{\partial}{\partial r}\right)+\dfrac{1}{r^{2}\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial}{\partial\theta}\right)+\dfrac{1}{r^{2}\sin^{2}\theta}\dfrac{\partial^{2}}{\partial\phi^{2}}$$

 

[PeroK]

Of course in general an operator can be a function of the coordinates; e.g., the 3D Laplacian in spherical coordinates:$$\nabla^{2}=\dfrac{1}{r^{2}}\dfrac{\partial}{\partial r}\left(r^{2}\dfrac{\partial}{\partial r}\right)+\dfrac{1}{r^{2}\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial}{\partial\theta}\right)+\dfrac{1}{r^{2}\sin^{2}\theta}\dfrac{\partial^{2}}{\partial\phi^{2}}$$

That's not a function of coordinates. It uses a coordinate function such as $r^2$ as part of its action on a function. The input to the Laplacian is a function and the output is another function.

 

[renormalize]

That's not a function of coordinates. It uses a coordinate function such as $r^2$ as part of its action on a function. The input to the Laplacian is a function and the output is another function.

OK, mea culpa! Yes, I was using "function" in the colloquial sense of "having a dependence on" the coordinates. Since a function by definition takes numbers into numbers, something taking functions into functions is indeed a different beast. Labeling it a "mapping", I modify my previous statement to simply note that this function-to-function "map" may well depend upon the coordinates.

 

[PeroK]

If $g(x,t) = f(-x,-t)$, then we could relate $\hat Ag$ to $\hat Af$.

 

[hokhani]

That's not a function of coordinates. It uses a coordinate function such as $r^2$ as part of its action on a function. The input to the Laplacian is a function and the output is another function.

In mathematical language, as you said, an operator is not a function of $(x,t)$. However, the operator depends on x and t. If we cannot write this dependence as $H(x,t)$, how to show this dependence in an operator equation?

 

[PeroK]

In mathematical language, as you said, an operator is not a function of $(x,t)$. However, the operator depends on x and t. If we cannot write this dependence as $H(x,t)$, how to show this dependence in an operator equation?

Can you be more specific about the context of the question?

 

[hokhani]

Can you be more specific about the context of the question?

Of course; in the Schrodinger equation we have something like: $H(x,t)\psi(x,t)=i\hbar \frac{\partial }{\partial t} \psi(x,t)$ which at each space-time point, $(x,t)$, only the Hamiltonian $H(x,t)$ acts on $\psi(x,t)$. I try to understand why $\psi(x,-t)$ is not the answer of $H(x,-t)$?

 

[PeroK]

Of course; in the Schrodinger equation we have something like: $H(x,t)\psi(x,t)=i\hbar \frac{\partial }{\partial t} \psi(x,t)$ which at each space-time point, $(x,t)$, only the Hamiltonian $H(x,t)$ acts on $\psi(x,t)$. I try to understand why $\psi(x,-t)$ is not the answer of $H(x,-t)$?

The Hamiltonian may depend on time, which means you do not have a simple operator equation. The Hamiltonian itself does not depend on position. The Hamiltonian is an operator. It's not a function.

Let's take the example of a time-independent Hamiltonian. This can be written in the form of an operator:
$$\hat H = -\frac{\hbar^2}{2m}\hat D^2 + \hat V$$We have used $\hat D$ to denote the differential operator and $\hat V$ to represent the operator generated by multiplication by some (potential) function $V$.

The action of this operator on a function, $\psi$ is:
$$\hat H\psi = -\frac{\hbar^2}{2m}\hat D^2\psi + \hat V\psi$$Finally, we can use $x$ and $t$ as the position and time variables for our functions and express this equation for every point $x$ and time $t$:
$$(\hat H\psi)(x, t) = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}(x, t) + V(x)\psi(x, t)$$If the potential is also a function of $t$, then we have a time-varying equation:
$$(\hat H\psi)(x, t) = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}(x, t) + V(x,t)\psi(x, t)$$In this case, you could write the Hamiltonian itself as a function of time:
$$\hat H(t) = -\frac{\hbar^2}{2m}\hat D^2 + \hat V(t)$$

 

[hokhani]

The Hamiltonian may depend on time, which means you do not have a simple operator equation. The Hamiltonian itself does not depend on position. The Hamiltonian is an operator. It's not a function.

Let's take the example of a time-independent Hamiltonian. This can be written in the form of an operator:
$$\hat H = -\frac{\hbar^2}{2m}\hat D^2 + \hat V$$We have used $\hat D$ to denote the differential operator and $\hat V$ to represent the operator generated by multiplication by some (potential) function $V$.

The action of this operator on a function, $\psi$ is:
$$\hat H\psi = -\frac{\hbar^2}{2m}\hat D^2\psi + \hat V\psi$$Finally, we can use $x$ and $t$ as the position and time variables for our functions and express this equation for every point $x$ and time $t$:
$$(\hat H\psi)(x, t) = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}(x, t) + V(x)\psi(x, t)$$If the potential is also a function of $t$, then we have a time-varying equation:
$$(\hat H\psi)(x, t) = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}(x, t) + V(x,t)\psi(x, t)$$In this case, you could write the Hamiltonian itself as a function of time:
$$\hat H(t) = -\frac{\hbar^2}{2m}\hat D^2 + \hat V(t)$$

Great answer, many thanks.

 

[hokhani]

Sorry for returning again to this question, I would like to be certain. So, I raise another question:
For the general form of the Hamiltonian in one dimension, $H=P^2/2m +V(x)$, which of the following is correct for $H\psi(-x)$?
$$ (1): \\
H\psi(-x)=(P^2/2m +V(x)) \psi(-x)$$ or $$ (2): \\
H\psi(-x)=(P^2/2m +V(-x)) \psi(-x)?$$

 

[PeroK]

The way to resolve such problems is to let:
$$\phi(x) \equiv \psi(-x)$$That's technically what $\psi(-x)$ means. That the wavefunction is a composition of the function $\psi$ with the function $x\to -x$. That is case (1).

However, you might also have the case of a general spatial inversion $x \ to -x$. That is case (2) where you have a general change of coordinates.