How to Choose which space to work in for Schrodingers equation?

Question

I'm working out of Shankar's principles of quantum mechanics book. And overall, I think I get the gist of how to solve problems with Schrodinger's Equation. I recall in my Modern Physics course, we only worked in the x-space it seems. Basically we limited our Schrodinger Equation to $$ \frac{d^{2} \psi}{dx^{2}} + \frac{2m}{\hbar^{2}} (E-V)\psi = 0 .$$ However, in Shankar he's introduced Bra-Ket notation and in familiarizing myself with it, I've run into trouble making sense of the exact path to go to solve a general problem.

For example, working with the free particle where $\hat{H} = \frac{\hat{P}^2}{2m}$. I follow so far, since $\hat{H} = T+V$ and $V=0$. I understand that the Schrodinger Equation becomes: $$i\hbar |\dot{\psi} \rangle= \hat{H}|\psi \rangle = \frac{\hat{P}^2}{2m}|\psi \rangle.$$

I still follow that the normal modes are solutions of the form $|\psi\rangle = |E\rangle e^{-\frac{iEt}{\hbar}}$ from the fact that when $\hat{H}$ acts on $|\psi \rangle$ it outputs an eigenstate $|E \rangle$ and we multiply that by the propgator which he derived for any arbitrary solution of Schrodinger's equation in the previous section (provided V has no time dependence). Once we put this back into the time-dependent Schrodinger Equation, which I've verified myself we get $$\hat{H}|E \rangle = \frac{\hat{P}^2}{2m}|E\rangle = E |E \rangle $$ The time-independent schrodinger equation. However, here is where he starts losing me. On the very next line he says the problem can be solved without going to any basis. And that any eigenstate of $\hat{P}$ is also an eigenstate of $\hat{P}^2$. Which I understand. What I don't understand is that the next like he trades out the $|E \rangle$ for $|p\rangle $ which to me, I think I need a little more justification for. I understand that when an operator acts on a state, it outputs an eigenstate. However, why are we in the $p$-space now if we don't need to go to any space to solve the problem?

He gets the allowed values for $p$, which depend on $E$. And then constructs the propagator for this problem.

I think my confusion comes from trying to make the leap from what I know to what this book is trying to show me. In Modern Physics, and in my undergrad quantum course, we were concerned with finding the wave function almost entirely. Once we found the wave function, and the allowed energies, we were done. What is the actual goal here? Constructing an orthonormal basis for Energy? Constructing the propogator? I apologize if my questions don't make much sense, I don't follow some of his work. And I have found some lecture notes online that illuminate it for me, but I think I still need clarification.

Also, in his particle in a box solution, he says he's solving it in the X basis and says its the only viable choice. Why is the P basis not a viable choice? In the well, in the p-basis, wouldn't the solution be simple like it is for the free-particle? I understand it complicates when there is a potential, because $\hat{H}$ doesn't only depend on $\hat{P}$ but I think I'm missing steps here as well.

I guess to break down my questions they are as follows:

1. In Shankar's free particle solution, what is the justification for going from a state that is in the E-space to the p-space?

2. What is the goal of solving the problem if not to get the wave function? We use some test state, $|E \rangle U(t)$ and then put it into the Schrodinger equation to get a solution for accepted values of p/E. Then solve for the propogator, somehow? I don't follow his steps.

3. How do you know which basis to work in for an arbitrary problem? Intuition? For the particle in the box he says the x-basis is the only viable one, but doesn't explain why. Should I try to work it out in the p-basis to justify his choice?

Answer 1

In Shankar's free particle solution, what is the justification for going from a state that is in the E-space to the p-space?

Shankar has skipped a crucial observational point in his calculation. First, he notes that $\mathbf{P}$ and $\mathbf{P}^2$ have the same eigenbasis, which is notationally written as $|p\rangle$. Now consider the time independent Schrödinger equation, which reads $$\frac{\mathbf{P}^2}{2m}|E\rangle = E |E\rangle.$$ What does this equation mean? This is just the eigenvalue equation for $\mathbf{P}^2$, and so it is literally telling us that $|E\rangle$ is an eigenstate of $\,\mathbf{P}^2$. Putting these observations together, we can simply replace $|E\rangle$ with $|p\rangle$, as Shankar does, and consequently we can treat $\mathbf{P}^2$ not as an operator but simply as a number $p^2$, which is the corresponding eigenvalue.

I will say, in my personal point of view, this calculation is not "basis independent," as Shankar claims. Indeed, we have literally used the momentum basis! Rather, it is a basis dependent calculation where we have never explicitly specified the form of the momentum basis in a more familiar basis (say, the position basis).

What is the goal of solving the problem if not to get the wave function? We use some test state, |⟩() and then put it into the Schrodinger equation to get a solution for accepted values of p/E. Then solve for the propogator, somehow? I don't follow his steps.

I'm not sure what you mean by a "test state" of the form $|E\rangle U(t)$, which is not a valid combination of an operator and a ket. I would say that there are two "goals" to the computation in Shankar. First is to find the eigenbasis and eigenvalues of the Hamiltonian. We have done this, namely they are the states $|p\rangle$ with eigenvalue $E = p^2/2m$. The second goal would then be to use this solution to solve for the time evolution of an arbitrary initial state $|\psi\rangle$. There are many (equivalent) ways of approaching this, but Shankar opts to use the propagator. I'm sure this derivation is somewhere in Shankar but let's review it. Generally speaking, if $\mathbf{H}$ is a time independent Hamiltonian, then the solution to the time dependent Schrödinger equation is $$|\psi(t)\rangle = \exp(-i\mathbf{H}t/\hbar)|\psi\rangle.$$ Define the time evolution operator $$U(t) \equiv \exp(-i\mathbf{H}t/\hbar)$$ Let $|E_\alpha\rangle$ be the eigenvectors of the Hamiltonian ($\alpha$ may be a discrete, continuous or mixed discrete-continuous index). Then, using the fact that the identity may be written as $$I = \sum_\alpha |E_\alpha\rangle \langle E_\alpha|,$$ we can express the right action of $U(t)$ as $$U(t) = \sum_\alpha \exp(-i\mathbf{H}t/\hbar) |E_\alpha\rangle \langle E_\alpha | =\sum_\alpha e^{-iE_\alpha t/\hbar} |E_\alpha\rangle \langle E_\alpha |.$$ This manipulation only tells us how the operator acts to the right, but an analogous computation should convince you that it also holds as a left action. Now for the free particle we should make the replacements $$\begin{align*}&E_\alpha \to p^2/2m \\ &|E_\alpha\rangle \to |p\rangle \\ &\sum_\alpha \to \int_{-\infty}^\infty dp \end{align*}$$ and so we obtain Shankar's expression in equation (5.1.9). The propagator tells us everything we need to know about time evolutions of states in terms of the initial state, so we have completely solved the Schrödinger equation.

How do you know which basis to work in for an arbitrary problem? Intuition? For the particle in the box he says the x-basis is the only viable one, but doesn't explain why. Should I try to work it out in the p-basis to justify his choice?

In finite dimensions, there is no "correct" answer to this question, and indeed, often times it is a matter of intuition to pick the most convenient basis. At the end of the day any basis will be as good as any other.

In infinite dimensional Hilbert spaces, such as for the free particle, square well or Harmonic oscillator, we must be a little more careful. In the example you give, as has been stated in other answers, the issue with naively trying to work in momentum space for the infinite square well is that the potential is not an integrable function, and so it does not have a Fourier transform. This is why Shankar says the position basis is the "only viable" basis. Shankar's claim is not strictly speaking true. We can in fact work with other bases, but we have to be careful about the space of functions we are working with. This point is a little abstract, and is the domain of a branch of mathematics called functional analysis, but we can be concrete in this specific example. Consider the infinite square well potential, $$V(x) = \begin{cases}0 & 0 < x < a\\ +\infty & \text{otherwise }\end{cases}.$$ Of course, this is not actually a well defined potential, since an ordinary real-valued function cannot be infinite anywhere on its domain. Really, what this is telling us is that we should restrict our space of functions from the set of all square integrable functions on the real line (called $L_2(\mathbb{R})$) to some smaller set of functions on a restricted domain. In this case, as we know, the set of functions we are interested in is all square integrable functions on the interval $[0, a]$ with the property $\psi(0) = \psi(a) = 0$. I will leave it to you to convince yourself that this is in fact a Hilbert space (with some appropriate googling and definitions). Now, what is a good basis on this Hilbert space? One is the position basis, as we are used to. Another is to use a discrete Fourier basis. For periodic functions on an interval, this basis has elements $$\{\cos(n\pi x/a)\}_{n=0}^\infty \hspace{1.5cm}\text{and}\hspace{1.5cm} \{\sin(n\pi x/a)\}_{n=1}^\infty$$ However, we are not done yet, because we have to satisfy the boundary conditions $\psi(0) = \psi(a) = 0$ The only basis elements which satisfy this condition are the $\sin$ functions, and so we conclude that a good basis for this problem is (up to a normalization) $$|n\rangle = \sin(n\pi x/a), n\geq 1.$$ As you may notice, this is the energy eigenbasis for the infinite square well Hamiltonian! So, if you had started in this basis, the Hamiltonian would already be diagonal and you would be done.

Answer 2

I did not look at your source so I can not tell you what exactly his goals are, but you seem to have a fundamental misconception of how an operator acts on a state. When you measure a physical quantity of a system, you will end up with a system, that is in an eigenstate of that quantities operator. Applying an operator to a state is not a measurement. If you apply an operator on one of its eigenstates, this will leave the state unchanged and just add (or rather multiply) the corresponding eigenvalue as a factor. If you apply an operator to a state that is no eigenstate, you will change the state, but you won’t end up in an eigenstate of the operator.

Now, what I assume is happening in your book, is that the author is using the fact, that you can write any state as a superposition of energy-eigenstates $|\psi\rangle = \sum_i E_i |i\rangle$, where $i$ does iterate over all possible eigenstates of the Hamiltionian. That means, that, if you expand your state like that, you can easily evaluate the Schrödinger equation: $$i\hbar \frac{\partial }{\partial t} |\psi \rangle= \hat{H}|\psi \rangle = \frac{\hat{P}^2}{2m}|\psi \rangle$$ (I think you forgot the $\frac{\partial }{\partial t}$ in your question) to calculate the propagator. However to get that representation you first need to solve the eigenvalue problem of the Energy-Operator (Hamiltonian) and that is your time-independent schrödinger equation $$\hat{H}|E \rangle = \frac{\hat{P}^2}{2m}|E\rangle = E |E \rangle $$.

Now concerning your confusion about the switching to $|p\rangle$: You already said that you understand that eigenstates of $\hat{p}$ are also eigenstates of $\hat{p^2}$. As your Hamiltonian is just a constant times the squared momentum $\hat{H} = \frac{\hat{P}^2}{2m}$, the same states will also be eigenstates of your Hamiltionian and therefore fullfill the time-independent schrödinger equation, which -as stated above- is just the eigenvalue problem of the Hamiltonian.

Now the motivation behind this is, I suppose, that the energy eigenstates will generally be degenerate, i.e. there will be multiple eigenstates for the same eigenvalue, which is, why I wrote $|\psi\rangle = \sum_i E_i |i\rangle$ and not $|\psi\rangle = \sum_E E |E\rangle$, which would be wrong, because you have to sum over more then 1 state per energy value. But one can write $|\psi\rangle = \sum_i E_i |i\rangle = \sum_{p_x, p_y, p_z} E( p_x, p_y, p_z) |p_x, p_y, p_z\rangle$, because the momentum eigenstates will not be degenerate as long as you neglect spin.