What does $f(x)$ satisfies the given equation means?

Question

In problem 2.1 part c of Introduction to Quantum Mechanics, 3rd ed. by Griffiths and Schroeter, they ask the reader to prove that if the potential is an even function of $x$, then if $\psi(x)$ satisfies the time-independent Schrödinger equation, then so does $\psi(-x)$.

Now the solution just requires one to prove that $\partial^2/\partial x^2=\partial^2/\partial (-x)^2$. Thereby the form of the equation does not change. All that changes is that the variable goes from $x\rightarrow-x$. So does the fact that the form of the equation does not change imply that $\psi(-x)$ is a solution?

Also, I belive that one can easily generalize this to time-dependent Schrödinger equation, but the fact that authors don't mention it makes me question the correctness of my claim.

Answer 1

As you note, the form of the equation doesn't change, which implies that $\psi(-x)$ is just as much a solution as $\psi(x)$ is. Similarly, one can take any linear combination $\phi(x)=c_1\psi(x)\pm c_2\psi(-x)$ and show that the form remains the same: $$-{\hbar^2\over 2m}{d^2\over dx^2}\phi(x)-V(x)\phi(x)=E\phi(x).$$ Whenever the time dependent Schrodinger equation can be solved by the technique of separation of variables, then if the time independent form admits $\psi(-x)$ as a solution then the time dependent form will also, so your intuition there is correct.

Now, whenever you are able to transform your solution and you find that the transformed solution is also a solution of the original equation, i.e. the form doesn't change, then the transformation is a symmetry of the Schrodinger equation for a given potential $V(x)$. In your example, the transformation is wrought by the linear transformation known as "parity". As a linear operator defined on the Hilbert space, the parity operator maps state vectors such that: $$\hat\Pi \;|\psi\rangle=|-\psi\rangle.$$ Because, as you have shown, parity is a symmetry of the Schrodinger equation for even potentials: $$\hat H\;(\hat\Pi|\psi\rangle)=E\;\hat\Pi|\psi\rangle,$$ thus $\hat\Pi\;(|\psi\rangle)$ is an eigenket of the Hamiltonian. Further, it is an easy exercise to show that if $|\psi\rangle$ is any even or odd eigen ket of the Hamiltonian then they are also eigenkets of the parity operator: such operators which share a common set of eigenkets are called compatible and are said to commute, i.e. $$[\hat H,\hat\Pi]=0.$$ This knowledge can be used to discover symmetries of the Schrodinger equation for some potential, for example, if one knows of an operator that commutes with the Hamiltonian, and further one knows the eigenkets of that operator, then one also knows automatically that those eigenkets are solutions to the Schrodinger equation. In your example, the $\psi(x)$, $\psi(-x)$ are eigenfuncitons of the parity operator which happens to commute with any Hamiltonian $\hat H=\hat p^2/2m+V(x)$, where $V(x)$ is even, thus $\psi(x)$ and $\psi(-x)$ are also eigenfunctions (solutions) of the time independent Schrodinger equation.

Answer 2

Now the solution just requires one to prove that $\partial^2/\partial x^2=\partial^2/\partial (-x)^2$.

Yes, if you can "just" prove the above-quoted equality you are basically "done." But have you understood what Griffiths hopes that you understand?

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It may be helpful to first think of an explicit example to see what is going on. Consider, for example, the (completely made up) function: $$ \psi_0(x) = x + (1+x)^2\;. $$ The function $\chi_0(x) = \psi_0(-x)$ can also be written down explicitly: $$ \chi_0(x) = -x + (1-x)^2\;. $$

Now look at the derivatives with respect to $x$: $$ \psi_0'(x) = 1 + 2(1+x) $$ and $$ \chi_0'(x) = -1 - 2(1-x) $$ and note that we have, explicitly in this specific case: $$ \chi_0'(x) = -\psi_0'(-x)\;. $$

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It is also true by the "chain rule" that, in general, whenever we have $$ \chi(x) = \psi(-x) $$ then $$ \chi'(x) = -\psi'(-x)\;. $$

By taking another derivative with respect to $x$ and using the chain rule we also see that, in general, whenever $$ \chi(x) = \psi(-x)\tag{1} $$ we have $$ \chi''(x) = \psi''(-x)\;. \tag{A} $$

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Thereby the form of the equation does not change. All that changes is that the variable goes from $x\rightarrow-x$. So does the fact that the form of the equation does not change imply that $\psi(-x)$ is a solution?

Thereby and forsooth, milord! However, I'm guessing that Griffiths probably wants you to be a little more specific in your proof... For example, first write down the starting point, which is that $\psi$ satisfies the TISE (I set $\hbar=m=1$ for my own sanity): $$ -\frac{1}{2}\psi''(x) + V(x)\psi(x) = E\psi(x)\tag{B} $$

Then use Eq. (A) above to see that: $$ -\frac{1}{2}\psi''(x) = -\frac{1}{2}\chi''(-x) $$ and, by Eq. (B), we see that this also equals $$ =E\psi(x) - V(x)\psi(x)\;. $$ and by Eq. (1) $$ =E\chi(-x) - V(x)\chi(-x) $$ and by the evenness of the potential $V(x)=V(-x)$ $$ =E\chi(-x) - V(-x)\chi(-x)\;. $$

Thusly and thereby and whence, we see that: $$ -\frac{1}{2}\chi''(-x) + V(-x)\chi(-x) = E\chi(-x)\tag{C}\;. $$

But the variable $-x$ in Eq. (C) can just be renamed to $x$ if you would like: $$ -\frac{1}{2}\chi''(x) + V(x)\chi(x) = E\chi(x)\tag{D}\;. $$

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Also, I belive that one can easily generalize this to time-dependent Schrödinger equation, but the fact that authors don't mention it makes me question the correctness of my claim.

Sure, yes, one can "easily" generalize this to the time-dependent Schrodinger equation. But, if it is so easy, then why are you questioning the correctness of the generalization? Anyways, the details of the generalization will be left as an exercise to the interested reader, since it is so easy. (Yes, it really is easy--or maybe better to say it is "straightforward".)