Treating the Schrödinger equation as an ordinary differential equation

How exactly can I interpret the equation in this form?

In order to develop a theory of solutions also in the general case, one interprets the time-dependent Schrödinger equation $$\mathrm{i}\frac{\mathrm d}{\mathrm dt}\psi(t)=H\psi(t)$$ as an ordinary differential equation for the vector-valued function $$\psi:\mathbb R_t\to\mathcal H := L^2(\mathbb R^{Nd})$$. However, since $$H$$ is not a bounded linear operator on $$\mathcal H$$, this theory turns out to be much more subtle than the theory of linear ODEs on finite dimensional spaces. These considerations will lead us to the spectral theorem for unbounded self-adjoint operators on Hilbert spaces, one of the central mathematical results discussed in this course.

So we treat $$\psi$$ as a curve in $$Nd$$-dimensional space, thus obtaining a solution for $$\psi(t)$$, but how could we ignore all the position dependence?

• What is $N$ and what is $d$? Also note that the Schrodinger equation doesn't necessarily involve position, is just usually viewed from the position basis, and so that is what we see most often. Oct 18, 2021 at 10:42
• $N$ the number of particles, $d$ the dimension of the space Oct 18, 2021 at 10:44
• I don't understand that very well, from my intuition, the wavefunction should depend on position, so I don't know.. is there maybe a reference I can read? Oct 18, 2021 at 10:53
• @khaled014z The position-space wavefunction depends on position, but e.g. the momentum-space wavefunction does not. Oct 18, 2021 at 11:14

$$\psi$$ is a curve through $$\mathcal H:=L^2(\mathbb R^{Nd})$$, not through $$\mathbb R^{Nd}$$ itself. That is, for each $$t\in \mathbb R$$ we have that $$\psi(t)\in L^2(\mathbb R^{Nd})$$ is loosely$$^\ddagger$$ a square-integrable function.

You must be careful to distinguish between $$\psi$$, which is a curve through $$\mathcal H$$, and $$\psi(t)$$, which is the element of $$\mathcal H$$ the curve passes through at time $$t$$. $$\psi$$ is a function of one variable which eats a time $$t$$ and spits out the vector $$\psi(t)$$, which is itself a square-integrable function. It wouldn't make any sense for $$\psi$$ itself to accept a position as an input - what would that even mean? Instead, it is $$\psi(t)$$ - which is often interpreted as a function of position (i.e. the position-space wavefunction) - which is able to take position as an input variable.

To make this explicit, we might use notation like $$\big[\psi(t)\big](\mathbf x)$$, which makes it clear that $$t\in \mathbb R$$ is a number which we plug into $$\psi$$ to get a function $$\psi(t)\in L^2(\mathbb R^{Nd})$$, and $$\mathbf x\in \mathbb R^{Nd}$$ is a vector we plug into $$\psi(t)$$ to get a number $$\big[\psi(t)\big](\mathbf x)\in \mathbb C$$. This notation is the stuff of nightmares, so I personally prefer the symbol $$\psi_t(\mathbf x)$$ instead.

In most of the pedagogical literature, authors tend to sweep this discussion under the rug and simply write $$\psi(t,\mathbf x)$$; however, my view is that this obscures the distinction between space and time which occurs in non-relativistic quantum mechanics, and leads to deep misunderstandings.

When we write down the Schrodinger equation, we need to interpret the terms carefully. In my preferred notation:

• $$\psi$$ is a vector-valued function of one variable, and $$\psi'$$ is its vector-valued derivative
• $$\psi_t$$ and $$\psi'_t$$ are the vectors in $$\mathcal H$$ which result from evaluating $$\psi$$ and $$\psi'$$ at time $$t$$
• $$H$$ is a linear operator which eats a vector and spits out another vector

$$\longrightarrow i \hbar\psi'_t = H\big(\psi_t\big)$$

$$^\ddagger$$Really an equivalence class of functions, where we identify two functions $$f$$ and $$g$$ as the same element of $$L^2(\mathbb R^{Nd})$$ if $$\int \mathrm d^{Nd}x |f(x)-g(x)|^2 = 0$$.

• Comments are not for extended discussion; this conversation has been moved to chat. Oct 19, 2021 at 16:36