Deriving the time-independent form of Schrödinger's equation

Question

The motion of particles is governed by Schrödinger's equation,

$$\dfrac{-\hbar^2}{2m} \nabla^2 \Psi + V \Psi = i \hbar \dfrac{\partial{\Psi}}{\partial{t}},$$

where $m$ is the particle's mass, $V$ is the potential energy operator, and $(-\hbar^2/2m) \nabla^2$ is the kinetic energy operator ($= p^2/2m$).

The state function can be expressed as the product of space-dependent and time-dependent factors, $\Psi(r, t) = \psi(r) w(t)$. If we substitute this into Shrödinger's equation, we get

$$\dfrac{-\hbar^2}{2m}w \nabla^2 \psi(r) + V \psi(r) w(t) = i \hbar \psi \dfrac{\partial{w}}{\partial{t}}$$

Upon dividing by $w(t)$, we get

$$\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r) = i \hbar \dfrac{\psi}{w(t)} \dfrac{\partial{w(t)}}{\partial{t}}.$$

But the time-independent Shrödinger equation is said to actually be

$$\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r) = E \psi.$$

I would greatly appreciate it if people would please take the time to explain what I did incorrectly here.

Answer 1

You didn't do anything incorrectly. You just haven't pushed far enough. You can write the last equation in your derivation as $$\frac{1}{\psi(r)}\bigg[\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r)\bigg] = i \hbar \frac{1}{w(t)} \dfrac{\partial{w(t)}}{\partial{t}}$$ Since the LHS and the RHS of this equation are independent of $t$ and $r$ respectively, they can be equal for some $\Psi(r,t)$ only if they are individually equal to a constant (think about it: otherwise, if I were to vary $r$ a little bit while keeping $t$ constant, the LHS would change but the RHS wouldn't and the equality wouldn't hold, so the conclusion is that each side should be equal to a constant, a constant that is independent of both $r$ and $t$). Let's call this constant $E$ (we will soon see that this is, in fact, the eigenvalue of the Hamiltonian, and thus, energy).

Thus, we say $$\frac{1}{\psi(r)}\bigg[\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r)\bigg] =E$$

This is the so-called time-independent Schr$\ddot{\text{o}}$dinger equation. As you can see, it is simply the eigenvalue equation for the Hamiltonian $-\frac{\hbar^2}{2m}\nabla^2+V$ and thus, the eigenvalue is rightly denoted by $E$!

Answer 2

As another answer points out, you didn't finish your separation of variables approach leading to the time independent Schrodinger equation (TISE).

However, there is another route to the TISE that might be of interest here.

First, the motivation for this problem is to find the wavefunctions that have definite values E of the (total) energy observable H. That is, we want to find the wavefunctions $\psi_E(x,t)$ that satisfy

$$H\psi_E(x,t)=E\psi_E(x,t)$$

where (working in 1D for simplicity here)

$$H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)$$

But the time dependent Schrodinger equation (TDSE) is

$$H\psi(x,t)=i\hbar\frac{\partial}{\partial t}\psi(x,t)$$

so the wavefunctions we seek satisfy

$$\frac{\partial}{\partial t}\psi_E(x,t)=-i\frac{E}{\hbar}\psi_E(x,t)$$

and it's easy to see that these wavefunctions are of the form

$$\psi_E(x,t)=e^{-i\frac{E}{\hbar}t}\psi_E(x,0)$$

where $\psi_E(x,0)=\psi_E(x)$ is a function of $x$ only. Now, put this back into the TDSE. First, note that:

$$H\psi_E(x,t)=He^{-i\frac{E}{\hbar}t}\psi_E(x)=e^{-i\frac{E}{\hbar}t}H\psi_E(x)$$

and

$$i\hbar\frac{\partial}{\partial t}\psi_E(x,t)=i\hbar\left(-i\frac{E}{\hbar}\right)e^{-i\frac{E}{\hbar}t}\psi_E(x)=e^{-i\frac{E}{\hbar}t}E\psi_E(x)$$

thus, after cancelling the common factor $e^{-i\frac{E}{\hbar}t}$, we have the TISE:

$$H\psi_E(x)=E\psi_E(x)$$

In summary, the solutions $\psi_E(x)$ of the TISE are the spatially dependent part of the wavefunctions $\psi_E(x,t)$ that have definite values $E$ of the energy observable $H$.

Answer 3

Let's understand where the time independent and dependent Schrodinger equation comes from, as they are both sides of the same coin. Let's first set $\hbar = 1 $ for simplicity. Time evolution of a state actually has a specific form - we can evolve the state with what we call a Hamiltonian $H$ by attaching an exponential term $$ \Psi(x, t) = \psi (x) e ^{ - i H t} .$$ When we make states dependent on time below, we say we are working in the Schrodinger picture. It is easy to check that this satisfies the equation $$ i \frac{ \partial \Psi }{\partial t} = H \Psi .$$ This is where the time-dependent equation you know and love comes from, and as you may notice, as you have written above, our Hamiltonian is precisely $$ H = - \frac{ \hbar ^ 2 }{2m} \nabla ^ 2 + V ( x)$$ Now to answer your question. It's not at all obvious what we could even split states and separate them into their time and spatial components as above. However, in your context, we assume we can. The spatial part $\psi$ is thus called a stationary state since we assume it doesn't depend on time and we also make the assumption that this state is an eigenstate of the Hamiltonian. Crucially, our energy is defined to be the eigenvalue of the Hamiltonian acting on the stationary on its own. So, we get that $$ H \psi ( x ) = E \psi ( x ) = \left[ - \frac{ \hbar ^ 2 }{2m} \nabla ^ 2 + V ( x)\right] \psi(x). $$ This set of eigenstates are the states which end up being physically observable - they form a basis of all states and when we observe a system we observe precisely one of these energy eigenstates.