# I The energy term in Schroedinger equation

1. May 12, 2016

The energy in the time-independent Schroedinger equation H|Ψ>=E|Ψ> is the total energy of the system,which is E=kinetic energy +V where V is the potential energy.
But while solving problems in quantum mechanics,like scattering problems and others,they deal with it as it is just the K.E.
For example in the potential step problem, they take the cases where E>V and E<V, but isn't that wrong since the energy of the system is obviously greater than V and can never be less than V?? because E=K.E+V and in text books they meant by E the the energy term in the Schroedinger equation?
What goes wrong here?
thanks.

2. May 12, 2016

### Staff: Mentor

You are mixing up a few things here. First, the energy is actually obtained as the expectation value of the Hamiltonian, $\langle E \rangle = \langle \psi | \hat{H} | \psi \rangle$.

Second, what is meant by "E < V" is that the total energy is lower than the barrier height V, not the total potential energy V. It might be less ambiguous if you take it as $E < V_0$, with the potential energy given as $\hat{V} = V(x)$, with $V(x)=0$ almost everywhere, except for $V(x) = V_0$ for $a < x < b$.

Also, scattering problems usually consider the case of an initial plain wave of (kinetic) energy E, since it originates in a region where there is no potential.

3. May 12, 2016

but when the particle enters the region where V=V0 doesn't this height barrier becomes the total potential energy?
I mean isn't the V(x) in the schroedinger equation is just meant to be the the potential energy of the system?

Last edited: May 12, 2016
4. May 12, 2016

### Staff: Mentor

No, since energy is conserved. That is why you get tunneling: for a particle to go from the left side of the barrier to the right when E < V, it has to tunnel through the barrier.

No, it is the potential energy operator. To get the actual potential energy of the system, you need to calculate $E_\mathrm{pot} = \langle \psi | \hat{V} | \psi \rangle$, or equivalently in position representation
$$E_\mathrm{pot} = \int_{-\infty}^{\infty} \psi^*(x) V(x) \psi(x) \, dx$$

5. May 12, 2016

### Orodruin

Staff Emeritus
This is the energy expectation value for an arbitrary state. The time independent Schrödinger equation, which is mentioned in the OP, is just the eigenvector equation for the Hamiltonian and the E appearing in it is the energy eigenvalue of a particular eigenstate we wish to find.

That being said, an eigenstate of the Hamiltonian does not have a definite potential energy, it makes no sense to, as the OP does, talk about V(x) as the total potential of the system. The potential is an operator which is different from the Hamiltonian and generally does not share the same eigenstates. It also makes little sense to talk about the expectation values for the Hamiltonian or potential for the scattering states as they are not normalisable.

6. May 12, 2016

@DrClaude @Orodruin
does this mean that the energy term of the Schrodinger equation is just meant to be the kinetic energy of the particle?
but this operator is included in the Hamiltonian operator.

7. May 12, 2016

### Orodruin

Staff Emeritus
So what? The sum of two operators need not have the same eigenvectors as the operators themselves.

8. May 12, 2016

### Staff: Mentor

Not, it is potential + kinetic energy.

9. May 12, 2016

### blue_leaf77

Well, that's quantum mechanics. A region which is classically forbidden becomes allowed when you go to the quantum world. Classically, the total energy cannot be less than the potential energy, but in quantum mechanics you have the Schroedinger equation which lets you have
$$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x) + V \psi(x) = E\psi(x)$$
with $E<V$, which is classically forbidden.

10. May 13, 2016

Right.
Lets take the potential step example,where for x<0 V=0 and for X>0 V=V0.
If the particle is in the region where where V=V0, what is the potential of the particle in this region?
I know this, but my question was about the nature of V in the Schrodinger equation.

11. May 13, 2016

### Orodruin

Staff Emeritus
But in the scattering eigenstates, the particle is not in any particular region. You could find a state which is fully contained in that region but its energy expectation value (it is not going to be an energy eigenstate!) is going to be larger than V. The scattering states have infinite support.

You simply cannot talk about the potential energy of a scattering state, it is not an eigenstate of the potential energy operator.

12. May 13, 2016

still confused :(

Yes that is because there is a reflection and transmission probabilities.
I didn't get this part, if you can simplify it more: why it is not going to be an energy eigenstate if I'm sure that it is in this region
The V in the schrodinger equation, refers to what then?

13. May 13, 2016

### vanhees71

I do not understand your problem. The Hamilton operator consists of a term we call kinetic energy and a potential as in classical mechanics. If your have problems to understand the physical meaning of the Hamiltonian you should go back to classical mechanics, before you try to understand quantum mechanics. A solid foundation in classical physics is an indispensible prerequisite to have a chance to get an idea about the meaning of quantum theory! You need to learn about Hamilton's principle in the Hamiltonian formulation, Poisson brackets and symmetry principles related to it (Noether's theorem) before you go back to quantum mechanics!

14. May 13, 2016

### Orodruin

Staff Emeritus
If it is it is an eigenfunction of the potential energy operator, but not of the kinetic energy operator (whose eigenstates are $e^{ikx}$). Hence, it is not an eigenstate of the Hamiltonian.
If $V$ here is a constant everywhere, it is also forbidden in quantum mechanics. It is absolutely necessary that $E < V$ in some region, since the kinetic energy operator is positive (semi-)definite.

15. May 13, 2016

### Orodruin

Staff Emeritus
While I do not disagree, I have often seen students being taught quantum mechanics without the slightest idea about neither the Lagrangian or Hamiltonian formulations of classical mechanics. I do agree that understanding and realising that quantum mechanics is not just some ad hoc assumptions but actually a development from classical mechanics is crucial, but it is not how it is often taught.

16. May 13, 2016

### blue_leaf77

I should have specified more about $V$, it should be written as $V(x)$ where $V(x)$ is a step potential or barrier or anything which has a portion for some values of $x$ where $V(x)$ is higher than $E$.
I was trying to answer this question:
since you forbade $E<V$ in quantum mechanics. It's forbidden in CM but allowed in QM.

17. May 13, 2016

### vanhees71

@Orodruin: So, how would you introduce quantum theory? I've no clue, how you'd do that without some ideas about Hamiltonian mechanics, particularly Poisson brackets.

For me, it's the greatest challenge of physics teaching to introduce quantum theory. I've not taught QM 1 (introductory lecture on non-relativistic quantum mechanics) yet. So I've never been in the dilemma to think about it. There are several approaches in the literature. In my opinion there are some no-gos, and these are only very slowly eliminated from the curricula in high schools and at universities (at least in Germany):

One should not start with old quantum theory; it's misleading from the very beginning; particularly to claim that the photoelectric effect "proves the existence of photons" a la Einstein's famous paper of 1905. Even worse is to teach Bohr's model of the atom (which BTW cannot be taught without the Hamiltonian formalism of classical mechancis either!). There's a reason why we don't use old quantum theory anymore but the modern way a la Schrödinger, Heisenberg, Born, Jordan, and (most importantly) Dirac!

On the other hand, of course, you cannot just through the postulates in abstract Hilbert-space language to the students and then perform dry calculations of all kinds of applications, but you need some heuristics to get there. I think, a good way is to start with wave mechanics to motivate the Hilbert-space formalism in terms of $L^2$, introducing operators representing observables and Born's rule (minimal interpretation). Then you have a good basis to do the formalism in the representation free (Dirac) way using the modern form of the "correspondence principle" based on symmetry considerations, leading to the operator algebra of observables. To keep the things as simple as possible you can use the Heisenberg algebra, i.e., the position and momentum algebra based on the definition of momentum as generators of spatial translations. Then it's also very natural that the Hamiltonian is the generator for the time evolution (but this must be taken with a grain of salt since time necessarily is not an observable but a parameter in quantum theory). For this you need Poisson brackets from classical mechanics motivating the commutation relations. Even when using no symmetry arguments but canonical quantization (which one should do with some caveat, because it only works for Cartesian coordinates of position and momentum and I find it a bit less physically motivated than to base theoretical physics on symmetry principles from the very beginning).

Later you can introduce also aspects of QT going beyond classical mechanics, using advanced methods of group-representation theory on Hilbert spaces, e.g., the idea of spin by analyzing the Galileo group. I've done this with great success in teaching QM 2 some years ago. The students liked this approach very much, and I think it makes clear in a very convincing way, why non-relativistic quantum theory looks as it does, at least it's more convincing than just telling the formalism and then justify it just by arguing "that it works" ;-).

18. May 13, 2016

I know all of these and I have a solid background in classical mechanics. But the problem is that you

19. May 13, 2016

### RockyMarciano

I believe amjad-sh problem here is precisely that he is holding on too much to the classical foundations, so I don't think it helps to point him to the classical picture. Quantum mechanics departure point is precisely to allow what is forbidden classically, in this particular case with respect to the concepts of potential energy and the Hamiltonian. So amjad-sh, if you don't want to encounter apparent paradoxes between total, kinetic and potential energies from the classical standpoint you basically have to unlearn certain things about the classic Hamiltonian now that the energy operator doesn't commute neither with the potential(except for constant potential) nor the kinetic energy operators. This produces relations between energy and potential energy that are just not compatible with how they were connected in classical physics with a classical Hamiltonian. You just have found one of the many consequences of non-commuting operators.

20. May 14, 2016

### vanhees71

Well, quantum theory hinges on the commutation relations of the operators that represent observables. What's new is the probabilistic meaning of the notion of states. So you need the classical foundation to understand the observable algebra following from symmetry principles. Also in classical mechanics the potential describes forces, and the Hamiltonian "doesn't commute" with it, because you want to describe forces.