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David Griffiths states in 'Introduction to Quantum Mechanics':

  1. The general solution is a linear combination of separable solutions. As we're about to discover, the time-independent Schroedinger equation yields an infinite collection of solutions ($\psi_1(x)$, $\psi_2(x)$, $\psi_3(x)$,...), each with its associated value of the separation constant ($E_1$, $E_2$, $E_3$,...); thus there is a different wave function for each allowed energy: $$\Psi_1(x, y) = \psi_1(x)e^{-iE_1 t/\hbar},\quad \Psi_2(x, y) = \psi_2(x)e^{-iE_2 t/\hbar}, \ldots.$$ Now (as you can easily check for yourself) the (time-dependent) Schroedinger equation has the property that any linear combination of solutions is itself a solution. Once we have found the separable solutions, then, we can immediately construct a much more general solution, of the form $$\Psi(x, t) = \sum_{n = 1}^{\infty}c_n\psi_n(x)e^{-iE_n t/\hbar}$$

Now I want to check this for myself. As an example, I choose:

$$\Psi(x,t)=c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}$$

Inserting this expression into the time-dependent Schrödinger equation, I get:

$$E_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+E_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}=\hat{H}\Psi(x,t)$$

Although I see that this 'looks' like the time-independent Schrödinger equation, I struggle to show that this linear combination of solutions is itself a solution. Although if we denote the potential operating on $c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}$ as $V_1$ and the potential operating on $c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}$ as $V_2$, we get:

$$E_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+E_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}=\hat{H}\Psi(x,t)$$ $$E_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+E_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}=[\frac{\hat{p}^2}{2m}+V]\Psi(x,t)$$ $$E_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+E_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}=\frac{\hat{p}^2}{2m}\left[c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}\right]+V\left[c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}\right]$$ $$E_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+E_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}=\frac{\hat{p}^2}{2m}\left[c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}\right]+V_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+V_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}$$ $$E_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+E_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}=\hat{H}_1c_1\psi_1(x)\text{e}^{\frac{-iE_1t}{\hbar}}+\hat{H}_2c_2\psi_2(x)\text{e}^{\frac{-iE_2t}{\hbar}}$$

With $\hat{H}_1=\frac{\hat{p}^2}{2m}+V_1$ and $\hat{H}_2=\frac{\hat{p}^2}{2m}+V_2$. This looks like the time-independent Schrödinger equation for each solution separately. This derivation rests on claiming $V=V_1+V_2$, but is this allowed and if so, why? Also; how does this show that a linear combination of solutions is itself a solution to the time-dependent Schrödinger equation?

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  • $\begingroup$ Your mistake is in not understanding that it needs to be $V=V_1=V_2$ and $H_1=H_2=H$ in order for the superposition to work correctly. $\endgroup$ Commented Jul 4, 2023 at 16:56

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The only thing you really need to assume is that the Hamilton Operator is a linear operator.

An operator $\hat{H}$ is called "linear" if $$\hat{H} (\psi_1+\psi_2) = \hat{H} \psi_1 + \hat{H} \psi_2$$ and $$\hat{H} (\alpha \psi) = \alpha \hat{H} \psi$$ for any complex number $\alpha$.

All operators that correspond to observables in Quantum Mechanics have this property. (Recall that operators on finite-dimensional Hilbert Spaces can be written as Matrices. This wouldn't be possible if the operators weren't linear.) The time derivative on the left-hand side of the Schrödinger Equation is obviously also linear. Therefore, if $$i\hbar \frac{\partial}{\partial t} \psi_1 = \hat{H} \psi_1$$ and $$i\hbar \frac{\partial}{\partial t} \psi_2 = \hat{H} \psi_2$$ then $$i\hbar \frac{\partial}{\partial t} (\psi_1+\psi_2) = \hat{H} (\psi_1+\psi_2).$$

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  • $\begingroup$ Just to nitpick, note that for $\hat H$ to be linear you also need $\hat{H} a \psi = a \hat{H}\psi$ for a scalar (complex number) $a$. $\endgroup$
    – Andrew
    Commented Jul 4, 2023 at 17:08
  • $\begingroup$ Thank you for the answer. It was very helpful. To me it seems natural that $\hat{H}a\psi=a\hat{H}\psi$. If a is just a scalar, shouldn't always be able to switch the scalar and the operator around? In what case, for example, does this expression not hold? $\endgroup$ Commented Jul 4, 2023 at 17:23
  • $\begingroup$ @RasmusAndersen We could define an operator that squares the function it acts on: $\hat{O} \psi := \psi^2$. This would be a non-linear operator. In that case, $\hat{O} (a \psi) = (a \psi)^2 = a^2\psi^2$, but $a\hat{O} \psi = a\psi^2$. $\endgroup$ Commented Jul 4, 2023 at 18:35
  • $\begingroup$ That is a very good point. Thank you. $\endgroup$ Commented Jul 4, 2023 at 19:33
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    $\begingroup$ Note that the fact that superposition works with the Schrödinger equation is because it is a linear differential equation. Any linear differential equation has this property, and that property makes it very easy to solve, as opposed to impossible for a lot of nonlinear differential equations. Thankfully, there's a whole lot of the physical world that can be modeled well enough with linear differential equations. $\endgroup$
    – TimWescott
    Commented Jul 4, 2023 at 20:24
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First, note that the time dependent Schrodinger equation is $$ i \hbar \frac{\partial \Psi}{\partial t} = \hat{H} \Psi $$

Let's suppose like in your example that $$ \Psi = c_1 \psi_1 e^{-i E_1 t / \hbar} + c_2 \psi_2 e^{-i E_2 t / \hbar} $$ Note that we can explicitly differentiate $\Psi$ with respect to time, yielding $$ i \hbar \frac{\partial \Psi}{\partial t} = c_1 E_1 \psi_1 e^{-i E_1 t / \hbar} + c_2 E_2 \psi_2 e^{-i E_2 t / \hbar} $$

Finally we can manipulate $\hat{H}\Psi$ as follows, using the linearity of $\hat{H}$ and $\hat{H}\psi_k e^{- i E_k t/\hbar}=E_k \psi_k e^{-i E_k t/\hbar}$ for $k=1, 2$. \begin{eqnarray} \hat H \Psi &=& \hat{H}\left[c_1 \psi_1 e^{-i E_1 t / \hbar} + c_2 \psi_2 e^{-i E_2 t / \hbar}\right] \\ &=& c_1 \hat{H} \psi_1 e^{-i E_1 t / \hbar} + c_2 \hat{H} \psi_2 e^{-i E_2 t / \hbar} \\ &=& c_1 E_1 \psi_1 e^{- i E_1 t / \hbar} + c_2 E_2 \psi_2 e^{- i E_2 t / \hbar} \\ &=& i \hbar \frac{\partial \Psi}{\partial t} \end{eqnarray} So the time dependent Schrodinger equation $i \hbar \frac{\partial \Psi}{\partial t} = \hat{H}\Psi$ holds when $\Psi$ is a linear combination of energy eigenfunctions.

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  • $\begingroup$ Thank you very much. I understand that $\hat{H}\psi_i=E_i\psi_i$ together with the linearity of $\hat{H}$ can be used to proove that TDSE holds for any linear combination of solutions. $\endgroup$ Commented Jul 4, 2023 at 17:26
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You only have to use the linearity of $H$ and $i\partial_t$.

On one hand, you have (using $H\phi_k(x)=E_k\phi_k(x)$)

$$ H\sum_{k}c_k\phi_k(x)\exp(-iE_kt)=\sum_{k}c_kH\phi_k(x)\exp(-iE_kt) = \sum_{k}E_kc_k\phi_k(x)\exp(-iE_kt) \ . $$ On the other hand, you also have

$$ i\partial_t\sum_{k}c_k\phi_k(x)\exp(-iE_kt)= \sum_{k}c_k\phi_k(x)i\partial_t\exp(-iE_kt)= \sum_{k}E_kc_k\phi_k(x)\exp(-iE_kt) \ . $$ These are equal, so $$ H\sum_{k}c_k\phi_k(x)\exp(-iE_kt) = i\partial_t\sum_{k}c_k\phi_k(x)\exp(-iE_kt) \ , $$ and thus the TDSE holds.

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  • $\begingroup$ Thank you for the explanation. $\endgroup$ Commented Jul 4, 2023 at 17:20

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