# Why is the first derivative of the time-dependent Schrödinger equation continuous? Where does it come from?

I was taught in first year physics that the first derivative of the time-dependent Schrödinger equation had to be continuous. However I was never taught (or at least I don't remember) the reason why.

$$$$\tag{1} i\hbar\frac{\partial}{\partial t}\Psi(\vec{x},t)=\left(\frac{\hbar^2}{2m}\nabla^2+V(\vec{x},t)\right)\Psi(\vec{x},t)$$$$

My suspicion is that it comes from the Schrödinger equation itself $$(1)$$. Perhaps some integral and derivative work will demonstrate that it's continuous for finite potentials?

At first I tried a proof from this page, thinking everything was fine and dandy, however I later realised that the proof was incorrect. It assumes the wrong definition for continuity, stating that a function is continuous at a point $$x_0$$ when $$f(x_{0}-\epsilon)-f(x_{0}+\epsilon)=0$$. This looks innocent enough, however if a function is discontinuous at only a point $$x_{0}$$, but not either side of it, then the definition will fail.

• Whay do think that the "assumption" is wrong? Feb 23 at 23:37
• Wavefunctions are only defined up to their value on sets of measure zero so the counter example you have in mind is in fact not a different state. Feb 23 at 23:40
• @mikestone Because of functions like this: i.imgur.com/ZyE2v9a.jpeg Feb 23 at 23:46
• @Jacob1729, can you please elaborate in an answer? Feb 23 at 23:46
• Possible duplicates: physics.stackexchange.com/q/262671/2451 , physics.stackexchange.com/q/19667/2451 and links therein. Feb 23 at 23:57

Wavefunctions in quantum mechanics are elements of the Hilbert space $$L^2(\mathbb{R})$$. This is usually described as the vector space of square-integrable functions with the inner product: $$\langle \psi,\phi \rangle = \int_\mathbb{R} \psi^*(x)\phi(x) dx$$

However one should note that one of the axioms for an inner product is that the inner product of any vector with itself is zero if and only if the vector itself is the zero vector. However, the above inner product has a non-trivial kernel (that is, functions $$\psi$$ such that $$\langle \psi,\psi\rangle=0$$.) This is precisely the set of functions that are zero almost everywhere but for a few isolated points where they have discontinuities. Thus, we actually construct $$L^2(\mathbb{R})$$ by 'quotienting out the kernel' - that is, we say two states $$\phi,\psi$$ are equivalent if their difference has zero norm:

$$\phi\sim \psi \iff \langle \phi-\psi , \phi-\psi \rangle = \int_\mathbb{R} |\phi-\psi|^2dx = 0$$

The states you have in mind that satisfy $$\lim_{x\to a^+}\psi = \lim_{x\to a^-}\psi \neq \psi(a)$$ are all equivalent under this equivalence relation to one where the value of $$\psi(a)$$ is simply swapped for the value of the left and right limits. Thus they aren't really different wavefunctions.

Physically what is happening is the wavefunction is used to construct expectation values via:

$$\langle Q \rangle_\psi = \int_\mathbb{R}\int_\mathbb{R} \psi^*(x)Q(x,x')\psi(x)dxdx'$$

and this integral simply doesn't care if you change the value of $$\psi$$ at only one point. So the answer to the question "are wavefunctions continuous" is really that they don't need to be, since you can add certain discontinuities that won't affect any observables. But the type that would affect integrals can be ruled out by the argument linked to in the OP.

• I like to avoid playing the devils advocate, but Dirac delta functions are discontinuous at only one point, and they have an area associated with them. So a wavefunction containing one of these will still affect observables. However, I think I understand the rest of what you are saying, and in practice we don't expect to encounter a wave-function that contains a Direc delta function. Feb 24 at 0:15
• @user400188 Dirac deltas are not functions, and to the extent that we can pretend they are, they are not square-integrable, so they cannot be featured in wavefunctions in the manner to which you refer. Feb 24 at 1:54