Consider the time-independent Schrödinger equation:

$$\operatorname{\hat H}\vert\Psi\rangle=E\vert\Psi\rangle$$

Is it not true that $E$ doesn't factor into any physically meaningful relation, and only $\Delta E$ does? That we can choose where we want $E = 0$.

Then, why is $E$ here? Where is $E = 0$ in this definition of $E$?

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    $\begingroup$ Consider redefining ${\hat H}$ by adding a term $u I$ to it, where $I$ is the identity operator. $\endgroup$ – Count Iblis Aug 26 '18 at 16:41

It is true that only $\Delta E$ matters and, therefore, it is possible to assign different values of $E$ to the same system without changing the meaning behind it. In other words, we can choose different reference points for the energy. However, the reference point is already well defined within the Hamiltonian.

For instance, take the example of a particle in a potential $V(x)=\frac{1}{2}kx^2$. The Hamiltonian will then be:

$$ H=\frac{p^{2}}{2m}+\frac{1}{2}kx^2 $$

Just like in classical physics, we could have defined the potential to be $V(x)=\frac{1}{2}kx^2+ \alpha$, where $\alpha$ is an arbitrary constant, and the physics would have been the same. However, the Hamiltonian now will be

$$ H=\frac{p^{2}}{2m}+\frac{1}{2}kx^2+\alpha $$

and, as a result, the original eigenvalues $E$ will become $E+\alpha$.

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    $\begingroup$ I've made some fixes to your mathematical notation - please see here for an in-depth tutorial. There's simply no need to use separate $...$ calls for different parts of the same formula. $\endgroup$ – Emilio Pisanty Aug 26 '18 at 17:11

Is it not true that $E$ doesn't factor into any physically meaningful relation, and only $\Delta E$ does?

Yes, that is accurate.

That we can choose where we want $E = 0$

Yes, that is accurate: if you want to shift all energies upwards by an energy change $\Delta$, the way you actually do this is by changing your hamiltonian from $\hat H_\mathrm{old}$ to $$ \hat H_\mathrm{new} = \hat H_\mathrm{old} + \Delta \, \mathbb I, $$ where $\mathbb I$ is the identity matrix (which is often dropped to avoid unnecessarily clunky notation, so you'd just see $\hat H_\mathrm{new} = \hat H_\mathrm{old} + \Delta$). Thus, if you have some eigenstate $|\Psi⟩$ of $\hat H_\mathrm{old}$ with eigenenergy $E$, then that same state also obeys $$ \left[\hat H_\mathrm{old} + \Delta \,\mathbb I\right] |\Psi⟩ = (E+\Delta)|\Psi⟩, $$ i.e. it is an eigenstate of $\hat H_\mathrm{new}$ with eigenvalue $E+\Delta$. It should be obvious that this transformation leaves all differences between eigenvalues intact, as required.

In other words, the hamiltonian is the energy in QM, period, and it is therefore, as an operator, as malleable with respect to the addition of an arbitrary constant offset as the classical energy.

  • $\begingroup$ The two statements are not entirely accurate. "the hamiltonian is the energy in QM" It is the non-relativistic approximation to the energy minus the rest energy. $\endgroup$ – my2cts Aug 26 '18 at 19:36
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    $\begingroup$ No, your comment is wrong. If you're doing relativistic quantum mechanics, then the hamiltonian also changes, and it is again equal to the energy. (On the other hand, if you're doing relativistic quantum mechanics, then both the energy and the hamiltonian are covariant components of a Lorentz four-vector, instead of an invariant like the Lagrangian; outside of a few well-controlled situations, you'll need to go to a full-blown quantum-field-theoretic treatment. But none of that changes the identification of the hamiltonian with the energy.) $\endgroup$ – Emilio Pisanty Aug 26 '18 at 20:39
  • $\begingroup$ I do not not say it is not, so your assertion is extremely misleading. I say that it is the non relativistic approximation to the energy minus the rest energy. If the non relativistic approximation to be accurate, it is the total energy minus the rest energy. $\endgroup$ – my2cts Aug 26 '18 at 20:53
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    $\begingroup$ No. You are thinking of the hamiltonian as $H=\frac1{2m}p^2+V$, and that is about as artificially narrow a view of hamiltonians as it gets. If you want to define $\hat H = +\sqrt{c^2 \hat p^2+m^2c^4}$ then that is also a perfectly valid hamiltonian for a free relativistic particle (and it is, moreover, the right hamiltonian) and it exactly coincides with the energy for that situation. $\endgroup$ – Emilio Pisanty Aug 26 '18 at 21:01
  • $\begingroup$ I don't recognize nor appreciate what you ascribe to me. This appears to be not about physics so I am signing off $\endgroup$ – my2cts Aug 26 '18 at 22:11

E is in the eigenvalue equation as a matter of definition.

Recall that in electrostatics, only differences in potential energy are meaningful. So this concept should not be too unfamiliar to you. It's just that in quantum mechanics we like to use the energy eigenbasis to express the quantum state of the system.

Is it not true that $E$ doesn't factor into any physically meaningful relation, and only $ΔE$ does?

This is true in general, but in the case that $|\psi\rangle$ is an energy eigenstate (NOT a superposition of energy eigenstates) then the expectation value of energy will only depend on the $E$ corresponding to that $|\psi\rangle$. It's the cross terms, or "interference" terms, that contribute the $\Delta E$.

As a specific example, consider the emission/absorption of light quanta: an atom in an initial energy eigenstate will lose energy by emitting a photon and end in an energy eigenstate of lower energy. Thus the photon (the thing we measure) will have an energy that is the difference of the two eigenstate energies.

Where is $E=0$ in this definition of E?

Typically, $E$ is nonzero due to the potential defined by the Hamiltonian (i.e. harmonic oscillator, hydrogen atom, infinte square well...) so this is all just a matter of definition. As Count Iblis said in the comment, if you just redefine your Hamilitonian as

$$ H' = H + uI $$

where $I$ is the identity matrix and $u$ is some nonzero constant, then you'll find that the energies $E'$ corresponding to $H'$ are necessarily nonzero.


Schrödinger energy is simply the total energy minus the rest energy in the low velocity approximation. One can include the rest energy and this will result in an additional high frequency phase factor $e^{imc^2t/\hbar}$. For energy transitions - energy differences in the same reference frame - this term is irrelevant. For the effect of a Galileo boost, $x'=x-vt, t'=y$ , this term has a large effect but this is outside the scope of the question.

Background: the Schrödinger equation is essentially the low momentum approximation to the relativistic Klein-Gordon equation*, which is the wave version of Einstein's energy-momentum relation, $E^2=m^2c^4+p^2c^2$. The zero order term, the rest energy, is left out. The

*$\partial_t\partial^t \psi = m^2c^4/\hbar^2\psi + c^2\vec \nabla \cdot \vec \nabla \psi$

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    $\begingroup$ (Not my downvote, but) this isn't really accurate. Your last statement is only true for $H={p^2\over 2m}$, which is far from the only valid Hamiltonian in QM. And it's not clear this really answers the question in any case. $\endgroup$ – Chris Aug 26 '18 at 22:52

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