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I was trying to learn Quantum physics by myself using MIT's 8.04 course. I came accross this equation:

enter image description here

I don't understand why the above is true. I understand the definition of linearity. But I don't know why two solutions to the above wouldn't be a solution if both sides involve derivatives (I am fully aware that one is a derivative wrt x and the other to t, but I don't know why that matters).

I don't understand why the above is not a linear equation and why adding/multiplying solutions doesn't work. Can someone give me an example?

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  • $\begingroup$ Forget about quantum mechanics, and consider a spring and mass system where the spring is nonlinear - for example the spring force is $-Kx - ax^3$. (That example is well known - google for the Duffing equation). $\endgroup$
    – alephzero
    Commented Aug 14, 2020 at 2:50
  • $\begingroup$ Welcome to Phys.SE. Avoid pasting images of text, type in the necessary content for your question in order to make the contents searchable.Use LaTeX/MathJax to typeset equations. $\endgroup$
    – ohneVal
    Commented Jan 21, 2021 at 8:38

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Solutions to the equation $x'(t) = x^2(t)$ take the form $$x(t) = \frac{-1}{t+C}$$

for some integration constant $C$. If I multiply $x(t)$ by some constant, does it remain a solution? What if I add two valid solutions together?

The problem is the nonlinearity on the right-hand side of the differential equation. If there were a linear function $V'(x(t))$ instead of the obviously-nonlinear quadratic function, then you can show quite easily that solutions can be added and multiplied by constants to yield other solutions. If the function is nonlinear, this is no longer true, as my simple example demonstrates.

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It's just that the solution of the differential equations are not in single powers of $t$. So, the equation is a non liner equation since the solution to the differential equation is non linear.

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    $\begingroup$ No, non-linear here means non-linear in $x(t)$. There are plenty of linear differential equations whose solutions are not linear, like $\dot x(t)=x(t)$. $\endgroup$
    – kaylimekay
    Commented Jan 21, 2021 at 8:20
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Great question and I know exactly what book you’re referring to. My answer is that if you want it to be linear then those two inequalities quoted in the text that you pasted in need to become equalities.

What does that imply? That taking the partial derivative of V(x) with respect to x when evaluated at x would have to be equal to the same when evaluated at ax.

Let’s think of some examples. Assume for a moment the derivative of the potential would simply be equal to x, the easiest of the potential polynomial representations. Well when you plug in X and aX they clearly wouldn’t be equal. So any polynomial wouldn’t work for V’. You can quickly surmise that other functions wouldn’t work either such as transcendental functions. As a result, the only ones that would work are those where the derivative would be a constant. In this way, evaluating it at different parameters changes nothing and it’s simply always the same constant so the equality would hold.

However the second condition regarding an operator on two solutions being the sum of the operator on each solution, where the operator in this case is V’, would also have to hold for linearity. And summing up a constant over two observations would never be equal to V’ evaluated once at the sum of the input parameters, it would always just lead to doubling, UNLESS the constant were zero.

In my opinion you would therefore need a potential that is constant so that the derivative of the potential, V’, is zero. Only in those cases would linearity be met in other words, both of the two inequalities would become equalities.

I hope this is helpful and I’m sorry if there are a lot of typos. I’m doing this all via dictation on my iPhone.

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The most general potential would be $V(x) = c_1 + c_2 x^2$. With $c_1$ and $c_2$ being constants. This is the potential for a harmonic oscillator. It is well known that the potential for a harmonic oscillator leads to a linear differential equation.

You can check that this potential gives a linear equation of motion by substituting $ax$ and $x_1+x_2$ in $V'(x)=2c_2 x$:

$V'(ax) = 2c_2 ax = a 2c_2 x =a V'(x)$

$V'(x_1+x_2)=2c_2(x_1+x_2)=2c_2 x_1 + 2c_2 x_2 = V'(x_1) + V'(x_2)$

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Interrogate the inequalities given as equations $(1.10)$. Imagine you're on a bike riding on a straightaway with rolling hills. When you're going uphill, there's a net force backwards on you (while you're acquiring potential energy) and when you're going downhill, there's a net force acting forwards on you (as you convert potential to kinetic). Now, is the net forward force on you generally proportional to how far you've gone? In other words, if there's a 3 Newton force acting against you at meter 3, was there also a 1 Newton force acting against you at meter 1? If it were, what would the profile of the hill you were on look like? If there was a 2 Newton force acting against you at meter 2 as well as at meter 4, would you expect a 4 Newton force acting against you at meter 6?

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