2:50 AM
morning

3:09 AM
morning

3 hours later…
6:02 AM
Morning.
Hello @JohnRennie Sir..

@123 hi :-)

@JohnRennie Does 1st and 2nd law of thermodynamics are completely independent?

@123 I'm a bit busy I'm afraid. I have a queue of people waiting to ask me questions this morning.

1st law tells us energy is conservative in thermodynamic system. Is there any real system which fulfill this requirement. Because every real system has dissipative force.
@JohnRennie Oooookay Sir.. Yes we are waiting for your guidance...

6:53 AM
Hello everyone

7:20 AM
What is an identity? The only way I can think of distinguishing two intellectual systems is by their unique observational standpoints in the environment (or a unique flow of observational standpoints). So if my brain gets cloned and put into a robot body, that robot is not me, since the moment it becomes active it starts to perceive different inputs, changing its internal structure. If on the other hand you cloned my biological brain and introduce exactly the same stimulus to me and my copy,
even tho the internal structure will preserve through time, the unique observational standpoint is the only way I can think of distinguishing these two entities and therefore calling myself the only me out of the two brains.
You can then say that since the observational standpoint changes through time, I'm a different person every moment, but what if we define identity as the unique flow (path) of observational standpoints through time?

7:36 AM
@JingleBells the ship of Theseus is an age-old argument about what identity means for divisible objects

8:01 AM
I'm a fan of the unique atom positions in space way to distinguish ships or people identities

4 hours later…
11:51 AM
Hello

12:12 PM
Can someone assist me with this technolgy?

hi, i was wondering if someone could assist me with queries related to quantum mechanics

When the hamiltonian is time independent, the resulting eigenfunctions will also be time independent(stationary states) and the linear combination of these eigenfunction will give a wavefunction which is general solution of the TDSE.
In a system with time dependent hamiltonian, will the corresponding eigenfunctions be also time dependent?
If yes would they still be called stationary states. Also incase the hamiltonian is time dependent will the general solution of the TDSE still be a linear combination of the eigenfunctions?

12:27 PM
@Physicsfreak we need to be a bit careful with terminology here
an eigenfunction is never time-dependent - it's just the eigenvector of some operator
when you have a time-dependent operator $A(t)$, you can talk about eigenfunctions of the operator $A(t_0)$ at a specific time $t_0$ but you can't talk about eigenfunctions of "the operator $A(t)$" for generic $t$ because it's not one operator, but a time-parametrized family of operators
the general solution to the TDSE is some function $\psi(x,t)$ which for time-independent Hamiltonians $H$ you can write as $\psi(x,t) = \sum_i \mathrm{e}^{-\mathrm{i}E_i t}\psi_{E_i}(x)$ where the $\psi_{E_i}(x)$ are eigenfunctions of $H$
when the Hamiltonian is time-dependent, you can still decompose any function $\psi(x)$ at an instant $t_0$ as the linear combination of eigenfunctions $\psi_{E_i,t_0}(x)$ of $H(t_0)$, but you don't get that the general solution to the TDSE has just these neat little phase factors $\mathrm{e}^{-\mathrm{i}E_it}$ as its time-dependence

@ACuriousMind so this means that its very complicated to find general solution of the SE when the hamiltonian is time dependent
5

I was reading through the proof of the Adiabatic Theorem (in Sakurai) and I realised I'm not quite sure how Schrodinger Basis kets behave when we have a time-dependent Hamiltonian. I know that with a time-independent Hamiltonian the basis kets don't change in the Schrodinger Picture. So if $|n;t... an answer to this question states that The Hamiltonian is just another operator. If the Hamiltonian is time-dependent, its eigenstates and eigenvalues are obviously time-dependent, too is it not correct? I'm sure Lubos means what I've written above, I just don't like his terminology He's saying that it depends on time which states are eigenstates of "the Hamiltonian" at that point in time I wouldn't say this means that "the eigenstates" are "time-dependent" because one might think we somehow have one operator where the eigenfunctions have magically become functions of time, this isn't what's happening @ACuriousMind so that means that the hamiltonian would have different eigenfunctions at different point of time? yes at each point in time, the operator$H(t_0)$has some set of eigenfunctions$\psi_{E_i,t_0}(x)$you could try to make a "time-dependent eigenstate" out of this via$\psi_{E_i}(x,t) = \psi_{E_i,t}(x)$, but this doesn't work: No one guarantees you that the Hamiltonian has the same eigenvalues$E_i$at each$t$(in fact, it might even switch from discrete to continuous spectrum or vice versa at some point), and this$\psi_{E_i}(x,t)$wouldn't be a solution of the TDSE, so the notation is misleading. so when the hamiltonian is time dependent, the linear combination of the eigenfunctions won't give a solution to the TDSE 12:42 PM do you mean$\psi(x,t) = \sum_i c_i \mathrm{e}^{-\mathrm{i}E_i t}\psi_{E_i}(x)$by "linear combination of the eigenfunctions"? I wouldn't really call that a "linear combination" since the coefficients are time-dependent, but to answer the question: No, that's not a solution for time-dependent Hamiltonians time-dependent Hamiltonians are very annoying and we usually try to avoid them where-ever possible @ACuriousMind thank you for the answer i have another question i hope you wouldn't mind answering it? 25 mins ago, by Nihar Karve > Don't ask about asking, just ask for a quantum system we never observe the wavefunctions, what we observe are the eigenvalues of the eigenfunctions. then why do we bother about getting a general solution to the TDSE? 12:48 PM because the wavefunctions are the means by which we obtain predictions for what eigenvalues we will measure with what probability @ACuriousMind don't we know what we are measuring in the system through the eigenfunctions? or through the operators I'm not sure what you mean like when we use the hamiltonian we know thatthe eigenvalues measure the energy levels, why do we need wavefunction to predict what are we measuring? @Physicsfreak but when you measure e.g. energy, you're not only interested in what eigenvalues you can measure, but also with what probability you will measure them okay so wavefunctions tell us what is the probabilty of say an electron to be in the ground state or the excited state? 12:52 PM The TDSE answers the following question: Suppose I start with my particle in a known state$\psi_0(x)$at time$t=0$. I wait a minute (time$t_1$) and then measure its position. The probability density that tells me how likely I will measure which positions is$\lvert \psi(x,t_1)\rvert^2$, where$\psi(x,t)$is the solution to the TDSE with the initial condition$\psi(x,0) = \psi_0(x)$@Physicsfreak yes, that's the Born rule @ACuriousMind is it correct to say that before we observe the quantum particle, its state is described by a wavefunction which is imaginary. as soon as we observe the particle, it collapes to one of the eigenstates of the wavefunction, the probability of collapsing to a certain eigenstate is determined by the probability density of the wavefunction the wavefunction is rarely imaginary, in fact you usually can choose it to be a purely real function @ACuriousMind even if its a purely real function , we can't measure it right? the rest is "correct", with the caveat that the exact ontological meaning of words like "probability" and "collapse" depends on your chose quantum interpretation because as soon as we measure it the system collapse to one of the eigenstates? 1:01 PM you can't directly measure the wavefunction, no @ACuriousMind that cleared my confusion. thanks for the help but the system is not necessarily in a unique state after measurement - operators can have more than one independent eigenstate for the same eigenvalue eigenspaces can be more than one-dimensional @ACuriousMind can you suggest where to read more about it? uhh...any text on linear algebra or QM? @ACuriousMind okay thanks 1:29 PM @Physicsfreak the idea is, if$i \hbar \frac{\partial }{\partial t} \psi(x,t) = \hat{H}(x,t) \psi(x,t)$is your partial differential equation, you have to solve it as it is, if$\hat{H}(x,t) = \hat{H}(x)$then we can try to mimic the method of 'separation of variables' and set$\psi(x,t) = \psi(x) \phi(t)$. Plugging this stuff in,$i \hbar \frac{\partial }{\partial t} [\psi(x) \phi(t)] = \hat{H}(x) \psi(x) \phi(t)$, we find$i \hbar [\frac{\partial }{\partial t} \phi(t)]/\phi(t) = \hat{H}(x) \psi(x)/\psi(x)$. The LHS is a function of time$t$, the RHS is a function of$x$(only because$\hat{H}$is time-independent!), so they are both equal to a constant$E$. We can solve$i \hbar [\frac{\partial }{\partial t} \phi(t)] = E \phi(t)$easily as$\phi(t) = e^{-iEt/\hbar}$, leaving$\hat{H}(x) \psi(x) = E \psi(x)\$, the TISE.

@bolbteppa, thanks for your answer but i wanted to know how to obtain the general solution when the hamiltonian is time dependent, what happens to the eigenfunctions of the hamiltonian. do you have any thing to add to this ?

@Physicsfreak if you're looking for a linear algebra textbook i'd recommend this
or if you want a qm book i'd recommend shankar

@Physicsfreak it can't be solved in general which is why people do perturbation theory

way too many QM books just drop new concepts out of the sky. but there are goods ones that smoothly transition to it from hamiltonian mech

@bolbteppa okay, we haven't covered that topic till now. thanks for your answer
@SirCumference thanks for the recommendation

1:43 PM
that's more than the total number of downvotes which I've ever received in my life (one of which was an obvious revenge downvote)
who even knows why

@NiharKarve it's rough if they don't give an explanation
best is to keep doing what you're doing till someone gives you advice, at which point you take it

@NiharKarve that's a normal amount of downvotes for HNQ answers that run counter to common pop-sci explanations :P

seriously
what is the motive there

I don't know since I don't know who is doing the downvoting
I just know that my popular answers that go in a similar direction also usually get downvotes in that range

1:59 PM
even your central extensions question has some downvotes, for no good reason whatsoever
oh well

@NiharKarve in that case curiously only the question, not the answer :P

it's bizzare

2:16 PM
Why would I need the temperature of the water when it's cold and warm in a heat pump cycle? Wouldn't I only need the refrigerant?

2:39 PM
In a condenser and evaporator, how would I account for in my thermodynamic analysis if their pressure are gauge pressures. Do I convert them? If so, how is that done, I feel like I might've done it incorrectly.

3 hours later…
5:29 PM
@Qwin You need the atmospheric pressure or elevation at the location of interest.

5:52 PM
Hello

6:14 PM