# Meaning of the term $V(x)$ in the Schrodinger equation [closed]

I'm new to quantum mechanics and I am currently trying to understand finite potential well (although my question is not specific to finite potential well ). In the Schrodinger equation, many texts and internet sources say that $$V(x)$$ is the potential. I have the following queries regarding this claim.

a) When they say $$V(x)$$ is the potential, is it just a way of "meaning" potential energy instead? Because after all, schrodinger equation is an energy conservation equation with kinetic and potential energies.

b) Secondly, if $$V(x)$$ means potential energy, then shouldn't the potential energy be dependent on the particle under consideration (I know a similar question has be asked before on this platform, but I am not convinced by the answer).

c) And if $$V(x)$$ is the potential energy, is it the potential energy that the particle under discussion would have had it been classically placed at the point $$x$$ or does $$V(x)$$ have the same meaning as it is traditionally defined, where a positive unit charge is placed at the point and it is the energy of that positively charged particle?

d) Lastly, when we are solving the finite potential well, we find the wave function for the two cases where $$E and $$E>V_0$$. Are these two cases for the same particle under discussion or are we saying "**if there is a particle whose energy $$E$$ is greater than $$V_0$$ (or if there is a particle whose energy $$E$$ is less than $$V_0$$) **

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• Potential energy = potential times charge. With this, the potential is only dependent on the properties of the source, not of the test particle. It's a convenient way of defining test particles. – AtmosphericPrisonEscape Jan 28 at 23:07
• There already some good answers. I just want to add a comment on c). $\textbf{Quantization is not well-defined}$ (I don't know enough about groupoids hahaha). You cannot expect to obtain the quantum theory from the classical one since the former is $\textit{more fundamental}$. This is the reason for your confusion. For example, in particles with spin in a magnetic field, there will be terms in the potential with no classical analog. Those you have to $\textit{learn by heart}$. However, when first learning quantum mechanics, we use classical potentials to build some intuition. – Iván Mauricio Burbano Jan 29 at 0:46

Yes, calling $$V(x)$$ the 'potential' is just a shorthand to 'potential energy'. And $$V$$ may depend on the particle we're investigating: if $$V$$ is the potential energy due to an electrical interaction then it must depend on the particle's charge. It's just that the potential well is a case where we may abstract away any dependence on the particle's properties. That's the answer to a) and b).

Now about c): we don't know how the interaction goes in this scale. We often 'discover' the appropriate $$V$$ by experiments or by looking at a similar classical interaction. If there's a similar classical case, we just take the classical hamiltonian and replace the $$x$$'s and $$p$$'s by the quantum mechanical operators $$X$$ and $$P$$ (being careful to keep the QM hamiltonian hermitian). In a nutshell: you can think of $$V(x)$$ as the potential a classical particle would have at $$x$$ only if $$V$$ has a classical counterpart, otherwise it's better to think that potential is set up in space and the particle responds to it at every position (since, anyway, a particle doesn't have a definite position in QM).

And for d): usually we treat them as separate experiments (where in one of them we used a particle with energy smaller than the 'potential wall' and in the other a particle with higher energy). But the most general solution for a QM particle is a linear combination of it's eigenfunctions, so you could have a particle prepared in such a way that both the $$E < V_0$$ and $$E> V_0$$ solutions make up the particle's state (an example: a particle has $$E, but we now make the particle interact with a photon; the particle may have absorbed the photon and increased it's energy or didn't - we can't really know before measuring! The best we can say is talk about probabilities and build the final wavefunction as a linear combination of states).

• Thank you for the answer. I just have a small clarification. So, should V(x) be considered as a potential energy operator or just the potential energy?. Because if it is the former, then your answer in part c makes even more sense to me than it already had. Because if it is an operator, then it would be silly to ask the question " what is the potential energy V(x) of the particle at some point x ? (Because the operator has not yet operated on the wavefunction yet to give us an answer). – Baikadi Pranay Jan 29 at 1:56
• In the most general case it is an operator which is a function of the position operator (so, as you said, if I don't apply it on a position eigenket it'll never make sense to say $V$ at point $x$). – ErickShock Jan 29 at 2:16

Let's see.

a) You are right, $$V(x)$$ is potential energy. It's a usual notation to use $$T$$ for kinetic energy and $$V$$ for potential energy. I cannot be more against this notation, because $$T$$ is period and $$V$$ is potential. We all know that, yet some teachers are able to mix both notations in the same blackboard without realizing it does not make any sense. The chaos is huge in the first times haha. I'm a rebel and I write $$E_p(x)$$, people look at me weird haha. But you'd be surprised about how many people re-discovers Schrödinger equation, like "Hey, you're right, that's potential energy, I mean, I knew it, but I had never realized so clearly", and all that. In short, we cannot swim against the stream so much, so I also use $$V(x)$$ sometimes, but be careful about what it means.

b) It is potential energy, and yes, it does depend on the particle under consideration. For example, an harmonic oscillator has $$V(x)=E_p(x)=\frac{1}{2}m\omega^2 x^2$$, and that depends on the mass of the particle, $$m$$. Or a Coulomb's potential (energy), which depends on the charge $$q$$ of the particle involved.

c) This is a delicate issue. QM works differently, and it cannot be explained in classical terms. The thing is that $$V(x)$$ has the same form as the classical potential energy, but replacing $$x$$ and $$p$$ by their corresponding operators.

That's the difference. It's the rule of quantization.

It cannot be explained classically, because you cannot place a quantum particle at one concrete point. That would be a Delta-funcion. So if position is not defined, you cannot apply the reasoning of work done the way it is classically.

However, you take the same form of the potential, and replace by operators. How you manage to achieve the conditions to create that potential is your problem haha. We just assume that the particle, for the fact of being there, is under the influence of such potential. And the way the potential affects the particle is ruled by the Schrödinger's equation.

d) A finite potential well (or potential barrier) has two cases, yes. On the one hand, the case where $$E>E_{p0}$$ and on the other hand the case where $$E. Those two cases are like two different problems. They are not mixed.

• Thank you for the answer, especially of parts c and d. Things are beginning to become a tad bit clear now, but I believe I still have a number of my own share of surprises to gather along this long road. I wanted to upvote your answer but I see that I dont have the necessary reputation. – Baikadi Pranay Jan 29 at 1:51