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...
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?
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?
@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
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...
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
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.
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
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?
@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
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 $\psi(x,0) = \psi_0(x)$
@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 rest is "correct", with the caveat that the exact ontological meaning of words like "probability" and "collapse" depends on your chose quantum interpretation
but the system is not necessarily in a unique state after measurement - operators can have more than one independent eigenstate for the same eigenvalue
@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 ?
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.
