it seems like all i can do is compute the expectation value from this, instead of the probability distribution

As an example conical pendulum. If we take origin A (center of the bob swing) here angular momentum constant in magnitude and direction and torque zero.

it would involve a change of basis back to the Schrodinger picture @ACuriousMind

but since the method is seemingly mass independent, that result would also be mass independent

@RyderRude I will not discuss the gravitational case with you until you've shown me you know how to actually do the derivation

If we take origin B (pivot where string is attached) angular momentum change in direction constant in magnitude. Also torque is constant in magnitude but change direction.

i dont know how....

but either way, i now dont think this method is mass independent

because of $X'(0)$

@123 I don't understand what the question is: Both torque and angular momentum involve a cross product with the position vector in their definition, so their values will in general change when you change the origin. This is a direct consequence of their definition. What exactly is the question?

@ACuriousMind do u genuinely want me to do the change of basis or r u just conveying that this method is shit?

I want to know how to see intuitively how to look angular momentum and torque at different origin. What i perceive from this. At origin A where torque zero. We see no force which causes angular velocity change. At origin B we see force which causes torque. Is it correct?

@RyderRude mostly I think you should do the derivation to get out of this habit of claiming to have "proofs" when you actually only have vague ideas and no actual derivation

ideas are good, but you need to know how to follow through on them

actually, i think i do have a derivation now

@123 I have no idea what "intuitively" means in this sentence

@ACuriousMind I understand and accept the consequence of definition. But what is the effect physically. So brain can see difference what are the meaning of changing magnitude and direction.

sorry, I really don't get the question

I look at $\vec L = \vec x \times \vec p$ and I think it's pretty obvious how this changes when $\vec x$ changes

so i got the final state as $\psi _i$ and $X(t)=x+\frac{P}{m}t$. now we use $X(t)=U(t) X U^{\dagger} (t)$. we do a change of basis using $U(t)$, which gives us $X(t)=x$ and $\psi (t)=U(t)|\psi\rangle$

and $U(t)|\psi\rangle$ gives us the cosine @ACuriousMind

What i think origin A where torque is zero, it means if i sit and see the conical pendulum from A it has no perpendicular force on bob because of zero torque. when i see conical pendulum at B it shows force on bob that's why we have torque constant magnitude. And torque means angular speed is changing.

@123 see if this helps you

@Amit Thanks

I think if i simulate conical pendulum using equations then i will find many answers. I learned a lot and found many answer by creating simulations.

But also learned here from all of you.

Physics is always served cold at the h bar

One issue I have with the site is the discouragement of check my work questions

"On 22 Dec 1988, the Plenum of the USSR Supreme Court ruled that Professor Fritz M. Noether had been convicted on groundless charges and voided his sentence, thus fully rehabilitating him."

how can i show that for a real potential, the time independent schrödinger equation has only real solutions as wave functions?

i am trying to build the complex conjugate and calculate with that but i am not reaching anything

@nickbros123 I think that the problem is that check my work questions necessarily have answers of the sort "You're wrong, but here's how to do it..." which encourages just asking homework questions in disguise. Or, will have answers of the sort "You're wrong/you're right/you're wrong, but here's a hint"... all of which are very short and are mostly of interest only to the one asking.

The only way to salvage such a question actually, is to address the physical concept underlying the question and thereby making it also of wider interest. In that sense I can agree -- maybe check-my-work questions should be allowed but the answers should be limited to only this kind

Because many times in the course of asking the question the OP reveals, unknowingly, his conceptual difficulty. That's just part of misunderstanding sometimes, you don't fully understand what exactly it is you don't understand ;-)

I'm floored by chatgpt's ability to make anologies: chat.openai.com/share/b6d234c2-c4ef-4f73-9a40-ce90ec8d035a

Today more anonymous learned mankind is doomed

Accordingly,i can congjuage and i have $d/dx^2 \psi^* = \gamma(x) \psi^*$

if i substract i have $d/dx^2 (\psi - \psi^*) = \gamma (\psi -\psi^*)$

not sure how i can conclude that the wave function is real

i know also theres a theorem that states that the solutions in this case are none degenerate

so i have $\psi = e^{i\phi}\psi^* $

@Amit oh, I was only talking about check my work questions that needs checking on the physical arguments one makes. Math anyone can do and those questions should be avoided. In the other case the clarification becomes a clarification on a concept, or a reason why one doesn't generally considers something the op has and so on, thus avoiding becoming a homework question and one that's actually useful

@nickbros123 If you're looking for clarifications on physical arguments I don't think it's check-my-work ;) sure, you can add some work to support your arguments, but it should be clear from the question that it's not the main thing you want to ask about

@user858770 Edward Fredkin passed away 3 days ago… apicciano.commons.gc.cuny.edu/2023/07/06/…

From en.wikipedia.org/wiki/Fredkin_gate

> The Fredkin gate (also CSWAP gate and conservative logic gate) is a computational circuit suitable for reversible computing, invented by Edward Fredkin. It is universal, which means that any logical or arithmetic operation can be constructed entirely of Fredkin gates. The Fredkin gate is a circuit or device with three inputs and three outputs that transmits the first bit unchanged and swaps the last two bits if, and only if, the first bit is 1.

@nickbros123 usually if one is in a position to ask to check their work, one also has an uneasy feeling that one or multiple steps they've done are possibly incorrect. so i think in most cases a check my work question can be translated into a conceptual misunderstanding question