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user87745

You didn't do anything incorrectly. You just haven't pushed far enough. Since $\psi=\psi(r)w(t)$ by stipulation, $i \hbar \dfrac{\psi}{w(t)} \dfrac{\partial{w(t)}}{\partial{t}}=i \hbar \psi(r) \dfrac{\partial{w(t)}}{\partial{t}}$. Thus, youYou can write the last equation in your derivation as $$\frac{1}{\psi(r)}\bigg[\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r)\bigg] = i \hbar \dfrac{\partial{w(t)}}{\partial{t}}$$$$\frac{1}{\psi(r)}\bigg[\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r)\bigg] = i \hbar \frac{1}{w(t)} \dfrac{\partial{w(t)}}{\partial{t}}$$ 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)$$\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 $E$!

You didn't do anything incorrectly. You just haven't pushed far enough. Since $\psi=\psi(r)w(t)$ by stipulation, $i \hbar \dfrac{\psi}{w(t)} \dfrac{\partial{w(t)}}{\partial{t}}=i \hbar \psi(r) \dfrac{\partial{w(t)}}{\partial{t}}$. Thus, you can write the last equation in your derivation as $$\frac{1}{\psi(r)}\bigg[\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r)\bigg] = i \hbar \dfrac{\partial{w(t)}}{\partial{t}}$$ 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 $E$!

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 $E$!

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user87745
user87745

You didn't do anything incorrectly. You just haven't pushed far enough. Since $\psi=\psi(r)w(t)$ by stipulation, $i \hbar \dfrac{\psi}{w(t)} \dfrac{\partial{w(t)}}{\partial{t}}=i \hbar \psi(r) \dfrac{\partial{w(t)}}{\partial{t}}$. Thus, you can write the last equation in your derivation as $$\frac{1}{\psi(r)}\bigg[\dfrac{-\hbar^2}{2m} \nabla^2 \psi(r) + V \psi(r)\bigg] = i \hbar \dfrac{\partial{w(t)}}{\partial{t}}$$ 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 $E$!