# Is the potential energy term of the Schrodinger equation correct?

The nonhomogeneous heat equation is of the form:

$$\frac{\partial }{\partial t} u(x,t) = \alpha^2 \frac{\partial^2}{\partial x^2} u(x,t) + f(x,t)$$

it appears as though we can always find some $g(x,t)$ and have written it with the substitution $f(x,t)=g(x,t)u(x,t)$ so that the solution $u(x,t)$ is unchanged, but the equation looks like this now:

$$\frac{\partial }{\partial t} u(x,t) = \alpha^2 \frac{\partial^2}{\partial x^2} u(x,t) + g(x,t)u(x,t)$$

The main difference (and the point of my upcoming question) is that $f(x,t)$ has a real life interpretation, it is an external heat source. On the other hand, $g(x,t)$ doesn't seem to really have any interpretation.

Now, when choosing a Hamiltonian it seems like there is a strong resemblance here. The potential energy term looks like $g(x,t)$ but for instance, the Coulomb force looks like $f(x,t)$. So intuitively it seems by analogy that the Schrodinger equation should really be in one dimension:

$$\frac{\partial }{\partial t} \Psi(x,t) = \alpha^2 \frac{\partial^2}{\partial x^2} \Psi(x,t) + V(x,t)$$

and NOT $$\frac{\partial }{\partial t} \Psi(x,t) = \alpha^2 \frac{\partial^2}{\partial x^2} \Psi(x,t) + V(x,t)\Psi(x,t)$$

so more specifically, in Dirac notation my question is:

Why is the Schrodinger equation $i \hbar \partial_t | \Psi \rangle = \hat T | \Psi \rangle + \hat V | \Psi \rangle$ instead of being $i \hbar \partial_t | \Psi \rangle = \hat T | \Psi \rangle + | V \rangle$? Alternatively I could have written in a way that is identical to my last statement but similar to the Schrodinger equation through use of the projection operator:

$$i \hbar \partial_t | \Psi \rangle = \hat T | \Psi \rangle + | \Psi \rangle \frac{\langle \Psi| V \rangle}{\langle \Psi | \Psi \rangle}$$

So to recap and try to make myself as clear as possible, $f(x,t)$ is a heat source and is analogous to the potential energy $| V \rangle$ term while on the other hand the meaningless $g(x,t)$ is analogous to the $\frac{\langle \Psi| V \rangle}{\langle \Psi | \Psi \rangle}$ which is the actual term found in the Schrodinger equation.

• you could try to work out the hydrogen atom with your alternative Schrödinger equation, and check whether it works or not. I pressume you won't get $E_n\propto 1/n^2$, which is the experimental result. In the end, it all boils down to matching the experiments. The eq. postulated by Schrödinger works very well, and yours (presumably) does not. – AccidentalFourierTransform Feb 3 '16 at 18:55
• Good idea, I'll try that. Already it gives the exact same result (unsurprisingly) as the free particle and particle in a box. I'm going to shoot for the SHO since that's 1D and see if I get anything that makes me interested in even trying the hydrogen atom. – user92640 Feb 3 '16 at 19:24

Quantum mechanics is a bit more than the Schrödinger equation. In particular, it says that all states evolve in time as given by the Schrödinger equation - there is the Hamiltonian operator $H$ and every time-dependent state $\lvert\psi(t)\rangle$ fulfills $\partial_t\lvert\psi(t)\rangle = -\mathrm{i}H\lvert\psi(t)\rangle$.

In contrast, the potential $V(x,t)$ is "intrinsically" time-dependent: It doesn't obey any evolution equation, in particular not the Schrödinger one. You cannot write $\lvert V\rangle$, it is not a member of the space of states, it is an operator on it. Aside from special cases (e.g. conformal field theory), there is no state associated to a generic operator. (To see why this is true, consider that for finite-dimensional vector spaces, the space itself has dimension $n$, but the space of operators has dimension $n^2$ - there is no way to map those nicely onto the states without losing a lot of information.)

• True, writing |V> was wrong to do, however writing it the final way I did is my true intention, essentially my point is that the potential energy can be written in the same eigenbasis expansion as |Psi> and instead of the operators imply blithely being "multiply by V(x,t)" the operator is the projection of V(x,t) into the eigenbasis of |Psi>. I'm trying to figure out how to rephrase my question since I am not really just simply "tripped up" by some technicality of the notation, I was trying to use the notation in a suggestive way that is incorrect. – user92640 Feb 3 '16 at 19:18
• @StevenGrigsby: You can't really expand $V$ in a basis in the same way as $\lvert\psi\rangle$ - because $\psi$ is a vector, while $V$ is the generalization of a matrix. Your final way seems to me to suffer from the same problem - $\langle\psi\vert V\rangle$ means the inner product of the vectors $\lvert\psi\rangle$ and $\lvert V\rangle$ by definition of the notation, but $V$ is not a vector. – ACuriousMind Feb 3 '16 at 19:36

The main point is that the wave function is postulated to contain the complete physical information of the system, in the sense that all there is to know is contained in the wavefunction. This is why Schrödinger's equation is an eigenvalue equation: this way the physical state at a later time is completely determined by the state at the current time:

$$i\hbar\frac{\partial \psi}{\partial t}=i\hbar\lim_{dt\to0}\frac{\psi(t+dt)-\psi(t)}{dt}=\hat H\psi(t).$$

If you write Schrödinger's equation like you do then the state at a later time is determined by the state at the current time and additionally by the potential at the current time:

$$i\hbar\frac{\partial \psi}{\partial t}=i\hbar\lim_{dt\to0}\frac{\psi(t+dt)-\psi(t)}{dt}=\hat T\psi(t)+V(t).$$

But this way, the equation for $\psi$ is not an eigenvalue equation any more, and this $\psi$ can not contain the complete physical information of the state, as postulated. Instead, the physical description of the state would only be complete if you add the "state" $|V\rangle$. (You can ask me now: "why is it postulated like this?". My answer: I have smart books on my bookshelf and they tell me it's this way. ^^ Besides, it's corroborated by experiments...)

Adding to this, since $\psi$ wouldn't contain all the information on the system, you would have problems with the interpretation of $$\left|\psi\right|^2.$$

So, normally the probability to find an electron in an interval $dx$ is $$\left|\psi(x)\right|^2 dx.$$ But in your picture, $\psi$ is incomplete, and you would need to incorporate the "state" $|V\rangle$ in some awkward way.