0
$\begingroup$

In the derivation of time-independent Schrödinger equation, after a certain point in time, LHS (time-dependent) and RHS (space dependent) are taken to be equal to a constant.

$$ i\hbar \frac{1}{\varphi}\frac{\partial\varphi}{\partial t} = -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{\partial^2\psi}{\partial x^2} + V(x) $$ Is this true for any case, say $x = 2t$ where $x$ and $t$ are any variables? Should both sides be equal to a constant?

$\endgroup$
  • $\begingroup$ It would be helpful to potential answerers if you included further details in your post. What are $x$ and $t$ in your example? Which exact step in the derivation don't you understand? (writing out the derivation of the time independent equation from the time dependent equation might help you pinpoint which step) $\endgroup$ – Samarth Nov 16 '17 at 9:34
  • $\begingroup$ I edited the question to typeset the equation in LaTeX. The site supports MathJax for formatting. You might find this helpful math.meta.stackexchange.com/q/5020 $\endgroup$ – Samarth Nov 18 '17 at 16:09
4
$\begingroup$

The derivation of the time-independent Schrödinger equation doesn't assume both sides equal a constant. It begins with the assumption that the wavefunction can be written as a product of two functions: one depending on the coordinate and one on time.

Edit: Adding details of argument

I am assuming $\varphi$ and $\psi$ are equations of just time and just position, respectively. That is,

$$ \varphi = \varphi(t) \\ \psi = \psi(x) $$

We have

$$ i\hbar\frac{1}{\varphi(t)}\frac{d\varphi}{dt}\left(t\right) = -\frac{\hbar^2}{2m}\frac{1}{\psi(x)}\frac{d^2\psi}{dx}\left(x\right) + V(x) \tag{1} $$

Note that the LHS is a function of $t$ and the RHS is a function of $x$, and $t$ is independent from $x$. Assume both sides are not constant, and that equality holds for some $t_o, x_o$.

$$ i\hbar\frac{1}{\varphi(t_o)}\frac{d\varphi}{dt}\left(t_o\right) = -\frac{\hbar^2}{2m}\frac{1}{\psi(x_o)}\frac{d^2\psi}{dx}\left(x_o\right) + V(x_o) = A \tag{2} $$

Since $t$ and $x$ are independent, if equality holds for $t_o, x$ such that $x \neq x_o$,

$$ i\hbar\frac{1}{\varphi(t_o)}\frac{d\varphi}{dt}\left(t_o\right) = -\frac{\hbar^2}{2m}\frac{1}{\psi(x)}\frac{d^2\psi}{dx}\left(x\right) + V(x) \tag{3} $$

(2) and (3) imply

$$ -\frac{\hbar^2}{2m}\frac{1}{\psi(x)}\frac{d^2\psi}{dx}\left(x\right) + V(x) = A \hspace{2em} \forall x $$

By similarly writing the equation for $t, x_o$ one gets the LHS of (1) to equal $A$ as well.

$\endgroup$
0
$\begingroup$

The equality $$ i\hbar\frac{1}{\varphi}\frac{\partial\varphi}{\partial t}= -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{\partial^2\psi}{\partial x^2}+V \tag{1} $$ must hold true for all times and all positions.

Thus, pick your favourite value for $t$ and call it $t_0$. This can be anything: $1,-33/7,\sqrt{\pi}$, whatever. Evaluate the left hand side of (1) at this $t_0$ so it becomes a number, which we call $\epsilon(t_0)$: $$ \left(i\hbar\frac{1}{\varphi}\frac{\partial\varphi}{\partial t}\right)_{t=t_0}=\epsilon(t_0)= -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{\partial^2\psi}{\partial x^2}+V \tag{2}\, . $$ This number $\epsilon$ may depend on $t_0$, but it's the same number for all $x$'s since the right hand side does not depend on $t$.

Now, start with (2) and note that (1) is still valid, so we can write $$ i\hbar\frac{1}{\varphi}\frac{\partial\varphi}{\partial t}= \epsilon(t_0)= -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{\partial^2\psi}{\partial x^2}+V \tag{3} $$ Pick now any fixed $x=x_0$ to evaluate rightmost term: $$ i\hbar\frac{1}{\varphi}\frac{\partial\varphi}{\partial t}= \epsilon(t_0)= \left(-\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{\partial^2\psi}{\partial x^2}+V \right)_{x=x_0} \tag{4} $$ The right hand side of (4) is a number which does not depend on $t_0$ since I can choose $x_0$ completely independently from $t_0$, yet the equality must still hold. Thus, $\epsilon$ does not depend on $t_0$ or $x_0$ and must be a constant: $$ i\hbar\frac{1}{\varphi}\frac{\partial\varphi}{\partial t}= \epsilon= \left(-\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{\partial^2\psi}{\partial x^2}+V \right)_{x=x_0} \tag{5} $$ Finally, use (1) again to get $$ i\hbar\frac{1}{\varphi}\frac{\partial\varphi}{\partial t}= \epsilon= -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{\partial^2\psi}{\partial x^2}+V \tag{5} $$ since the left hand side of (1) is always equal to its right hand side.

In other words, it is because (1) is valid not just for $x=2t$ for but any $x$ and any $t$ that one eventually reaches (5). Moreover, both sides of (1) must be equal to the same constant $\epsilon$, rather than each side equal to its own constant.

$\endgroup$
0
$\begingroup$

This will be true for any case where x and t are independant of each other. Because ,suppose you change the value of $t$. Let t be changed from $t_1$ to $t_2$ But for any $t$ we have $$x=2t$$ Then $$x=2t_1$$ and $$x=2t_2$$ Since $x$ is not varying the right hand side of these equation must be equal. This implies that $$2t_1=2t_2$$ This gives $$t_1=t_2$$ Which means $t$ is not varying hence $t$ must be a constant let $t=k^*$. Similarly, if you write down equations for $x_1$ and $x_2$ you will get $x_1=x_2$.which again gives $x$ is also a constant. Let $x=k'$ So $2t=2k^*=H$ again a constant. But we have $$2t=x$$ that is $$k'=H$$ and both side is equal to the constant which equals: $$k'=H$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.