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I am new to QM but I don't really understand when to use time dependent or time independent schrodingers equation? For example, for a free particle in a box, how do we know we should use TISE. more specifically, how do you determine if a wave function should be time dependent?

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  • $\begingroup$ The answers below are great, but they don't explicitly address "how do you determine if a wave function should be time dependent?" I'm sure you see from @ravjotsk that the wave function always has time dependence, it's just that in using the TISE we know the time part is just an exponential function, so people tend to focus on the space part more than the time part. But never forget it is still there. $\endgroup$ – Aaron Stevens Jul 25 '18 at 10:32
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In general, you always need to solve the Time dependent Schrodinger's equation. However, when your Hamiltonian is not an explicit function of time, you can solve the Time Independent Schrodinger's equation first and get the time dependent result easily. The TDSE $$i\hbar \frac{\partial}{\partial t}\Psi (x,t) = \hat{H} \Psi (x,t) $$ when $\hat{H}$ does not explicitly depend on time, solution of the equation is given by $$\Psi (x,t) = e^{-\frac{i}{\bar{h}} \hat{H}t} \Psi(x,0)$$ Here $\hat{H}$ is an operator and $\Psi(x,0)$ is your initial state.

Note that if $\Psi(x,0)$ is an eigenstate of the $\hat{H}$ with eigenvalue $E_n$, then by expanding the exponential function as a series of operators and acting on the eigenstate, you get $$ e^{-\frac{i}{\bar{h}} \hat{H}t} \Psi(x,0) = e^{-\frac{i}{\bar{h}} E_n t} \Psi(x,0)$$

The idea is that if we know the initial wavefunction in terms of eigenstates, then we can get the time dependent solution very easily.

Now the eigenstates of a hermitian operator forms a complete orthonormal basis, meaning that any random state can be written as a linear combination of eigenstates. Suppose you know all the eigenstates of the hamiltonian. Let $\psi_n$ represents an eigenstate with eigenvalue $E_n$. Then $$\Psi (x,0) = \sum_n c_n \psi_n(x) $$ You can find the coefficients using the orthonormality condition. This gives $$ \Psi(x,t) = \sum_n c_n e^{-\frac{i}{\bar{h}} E_n t} \psi_n(x) $$

Hence all you need to do is solve the TISE $$\hat{H} \psi(x) = E \psi(x) $$ and you can get the general time dependent solution easily for any initial state.

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    $\begingroup$ Thanks for the great answer. Just to confirm, assuming the Hamiltonian is time-independent, it is hence possible to solve the Schrodinger's Equation for any initial condition and Hamiltonian right? $\endgroup$ – Chien Hao Tan Jul 25 '18 at 14:26
  • $\begingroup$ Yes, as long as you are able to solve the TISE. $\endgroup$ – ravjotsk Jul 28 '18 at 11:54
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Ok. This is a basic confusion for many who are being introduced to Quantum-Mechanics for the very first time. I shall try my best to address your question, qualitatively. But only for time-independent Hamiltonians.

In quantum-mechanics problems, you start with the TDSE, which is a partial differential equation. As a standard method in the theory of partial differential equations, the technique of separation of variables is employed to write the space-and-time dependent wave-function as a product of two separate functions: one of them being exclusively dependent on position alone, and the other on time.

Once this is done, the partial differential equation(i.e. TDSE) splits up into 2 ordinary differential equations: one for the space part, and the other for time.

The time part is easy to solve, and it is the space-part that gives you the eigenvalue equation- the so-called TISE.

Solving this eigenvalue equation is the most important. For it gives you the energy-eigenspectrum for the problem at hand, and once this is obtained, you can multiply the time-part to each of these energy-eigenstates to make them evolve in time. However, being "stationary states", each of these energy-eigenstates keeps on maintaining the same energy eigenvalue(since the Hamiltonian doesn't explicitly depend on time).

But since your question is quite simple enough, I am avoiding time-dependent Hamiltonians for now, and to get to the heart of your question, my answer is this:

The quantum-mechanical wave-function for any given system is intrinsically time-dependent. The equation for it is given by TDSE. It is after using the separation of variables technique that we get two ordinary differential equations, out of which, the space-dependent one gives us the TISE, where one deals with a function(don't call it the "wave-function", it's misleading), that depends only on the position-variable(s).

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  • $\begingroup$ Thanks alot! Just a further question, what if the wave function is not separable in time and space? Do we need to employ a different approach, or can we use the result from ravjotsk's response to conclude that any wave function can be represented as a linear representation of separable eigenstates (and hence the SE is solvable for any initial wave function condition)? $\endgroup$ – Chien Hao Tan Jul 25 '18 at 14:27
  • $\begingroup$ That's exactly the whole point of it: one simply can't decompose the wave-function into a time and a space-part, unless and until the Hamiltonian involved is explicitly time-independent. Maybe you would like to go through this response as well: quora.com/… $\endgroup$ – Sudeepan Datta Jul 25 '18 at 16:05

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