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I am a bit confused with the Schrödinger equation. It describes the state of a system, the system could be of any size (although the equation will be hard to solve) but temperature doesn't play any role there.

So Why isn't temperature involved in the Schrödinger equation? Is it valid only a $T=0$?

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    $\begingroup$ How are you defining temperature? $\endgroup$ – tpg2114 Sep 29 '17 at 21:34
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    $\begingroup$ The equation deals with the energy of the system which includes thermal energy. $\endgroup$ – shai horowitz Sep 29 '17 at 21:36
  • $\begingroup$ something related to the movement respect to something? @tpg2114 I am not sure.. $\endgroup$ – user153036 Sep 29 '17 at 21:37
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    $\begingroup$ Why don't Newton's laws refer to temperature? $\endgroup$ – Rococo Sep 29 '17 at 23:32
  • $\begingroup$ because T doesn't affect movement of those bodies, which is not true for molecules..@Rococo any better comment? $\endgroup$ – user153036 Sep 29 '17 at 23:55
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Schrödinger's equation makes a microscopic-level description, while temperature is a macroscopic variable.

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  • $\begingroup$ My physics teacher told us that temperature is defined at a "mesoscopic scale". Too few particles : it's undefined, too many particles : the temperature isn't uniform enough to be correctly described by one single value. $\endgroup$ – Eric Duminil Sep 30 '17 at 14:02
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    $\begingroup$ @EricDuminil Yes, it's important not to try to apply temperature to anything larger than the 2.725K cosmic microwave background. ;-) $\endgroup$ – Xerxes Sep 30 '17 at 15:07
  • $\begingroup$ @Xerxes: Thanks for the laugh. It only works for the cosmic microwave background because it's surprisingly uniform. For large bodies, you can usually only talk about an average temperature or the temperature of a very specific location. Even for the CMB, it's "2.7260±0.0013 K", not just "2.7260K". $\endgroup$ – Eric Duminil Sep 30 '17 at 15:43
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    $\begingroup$ I think this answer is wrong. See the answers by Bruce Lee and my own to see why. $\endgroup$ – thermomagnetic condensed boson Mar 2 '18 at 18:08
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    $\begingroup$ So do I (and I assume Bruce Lee too), FGSUZ. $\endgroup$ – thermomagnetic condensed boson Mar 6 '18 at 16:57
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Temperature is a statistical property of an ensemble of particles/molecules. The Schrödinger equation works from the perspective of a single quantum state.

I'm sure that it might be possible to introduce temperature in one way or another to the Hamiltonian, but there is seldom a need for this when working with a single particle/atom in quantum mechanics.

As pointed out by DanielSank, the temperature of the environment that interacts with a system will very likely enter the Hamiltonian if it is relevant to the studied process.

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    $\begingroup$ "but there is seldom need for this when working with single particle/atom in QM" That's really not true in practice. $\endgroup$ – DanielSank Sep 29 '17 at 23:29
  • $\begingroup$ I'd love to correct my answer! Do you have any examples? $\endgroup$ – Darkseid Sep 29 '17 at 23:40
  • $\begingroup$ Well this may not have been what you had in mind when you said "single particle", but in real life nothing is ever completely isolated from its surroundings. If I have a single quantum system $S$ in contact with some other environmental degrees of freedom $E$, the temperature of $E$ determines the effect that they'll have on $S$. This is still referred to as "single particle" quantum mechanics because we're only calculating the dynamics of $S$ (even though $E$ is there and has an effect). $\endgroup$ – DanielSank Sep 29 '17 at 23:42
  • $\begingroup$ Edited answer to include mention of how temperature could enter Hamiltonian. $\endgroup$ – Darkseid Sep 29 '17 at 23:45
  • $\begingroup$ And what about when there is no environment, for a single molecule? Are we on T=0? Or it has no sense talk about T? $\endgroup$ – user153036 Sep 29 '17 at 23:57
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A thermodynamic state characterizes the thermodynamic observables of a system at equilibrium like temperature or pressure. By definition these observables are defined in the thermodynamic limit, i.e, when the number of degrees of freedom of the system are large. In other words, they describe the emergent macroscopic features of the system, rather than describing all possible microscopic features.

The thermodynamic variables can enter the Schrodinger equation or the Newton laws depending on the system you choose. Eg, if you take a particle in a box interacting with the environment then the relevant thermodynamic variables enter the description using the Hamiltonian for the particle, and thus enter both the Schrodinger equation as well as Newton's laws. Open quantum systems where degree(s) of freedom interact with environment may have temperature and other macroscopic observables characterizing the state of the environment entering into the Hamiltonian for the degree(s) of freedom.

Usually the Hamiltonians we consider in a QM1 course (I think that's where your confusion is coming from) don't have this kind of degree(s) of freedom-environment interaction, so thermodynamic variables don't enter the description.

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  • $\begingroup$ Although I don't understand it at all, this answer seems to me much better, Thanks! $\endgroup$ – user153036 Sep 30 '17 at 4:11
  • $\begingroup$ So, in quantum theory there is no temperature? Molecules are in some place without it? $\endgroup$ – user153036 Sep 30 '17 at 4:13
  • $\begingroup$ Thats not what I said, some specific Hamiltonians which you may have seen may not have any temperature dependence but in some cases they might have. $\endgroup$ – Bruce Lee Sep 30 '17 at 4:15
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    $\begingroup$ I think this answer is better than the currently accepted (and top voted as well) answer. It is better in that it is right (as opposed to wrong as the accepter answer is) and more detailed. See my own answer too for other details. $\endgroup$ – thermomagnetic condensed boson Mar 2 '18 at 18:14
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It may help to return to the classical mechanics of particles, whose exposition -- (see, e.g. Landau and Lifshitz, 'Mechanics') is altogether distinct from the exposition of thermodynamics -- in itself a separate discipline.

I would then make the analogy between classical and quantum mechanics, to assert that Schroedinger's equation is the quantum mechanical analog of the classical mechanics of particles, and therefore -- as in the classical case-- thermodynamic considerations, particularly that of temperature, do not enter.

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  • $\begingroup$ nice answer, but it is a bit strange anyway.. $\endgroup$ – user153036 Sep 29 '17 at 21:55
  • $\begingroup$ because..if ths system is big enough, then T should appear..At least seems logic to me.. $\endgroup$ – user153036 Sep 29 '17 at 21:58
  • $\begingroup$ Read my answer below. You can describe a chair by the position and velocities of all its forming particles... or you can use macroscopic variables such as temperature and give "more incomplete" but sufficient and useful information. $\endgroup$ – FGSUZ Sep 29 '17 at 22:06
  • $\begingroup$ @FGSUZ nice answer, but I need a bit more information..for example, would be the same a schrodinger equation for a big system at any temp? You say temperature appears in macroscopic treatment, but those theories need to be consistent..if T changes reality it must change thermodynamics results and Schrod results.. $\endgroup$ – user153036 Sep 29 '17 at 22:59
  • $\begingroup$ They are completely consistent. A chaneg in T means a change of Energy and that's what appears in either classical or quantum mechanics in microscopic treatements. If you have a microscopic system, you have to calculate the entropy and then T is defined so that $\frac{1}{T}=\left( \frac{\partial S}{\partial U}\right)$ $\endgroup$ – FGSUZ Sep 30 '17 at 9:45
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My take on this goes against the accepted answer's approach. One can use Schrödinger equation to model a macroscopic system, for example a crystal. Moreover, one can imagine systems in which the potential term that appears in Schrödinger equation depends on temperature. The result is that the Schrödinger equation involves temperature.

Consider a crystalline solid in which the interatomic distance depends on temperature. The Hamiltonian applied to the state of the system (which is nothing but Schrödinger equation) will involve terms that depend on temperature.

And to answer the last question of santimirandarp, while it is possible to model a system at $T=0K$, Schrödinger's equation is valid at any temperature.

Edit: As have been requested, in order to improve the answer, I respond to a comment by giving more precision, based on Ashcroft and Mermin's book "Solid State Physics", chapters 22 and 23 (only prerequisite is chapter 5 according to the book's plan) as a reference. It is a question of harmonic crystals. These are crystalline solids where the position of atoms or ions is not fixed but is allowed to vary in such a way that in (a time) average, their position is fixed and correspond to a Bravais lattice site. Furthermore, they may not move too far away from their equilibrium position, which is a reasonable assumption to describe many (but not all) properties for many, if not most, solids. This is called the harmonic approximation. As an example, oscillating ions about their equilibrium position, just like a mass on a spring, would satisfy this criteria. According to the book, by making a further approximation, called the adiabatic approximation (which is based on assuming that electrons move much faster than the ions of the solid, which is generally a justified approximation and the result is that the electrons are to be found in their ground state for that particular ionic configuration), statistical mechanics can now usefully enter into play. With the classical treatment, it is shown that the ionic displacement of an ion is worth $\vec u (\vec R) = \sqrt{k_\text{B}T} \vec u (\overline{\vec R})$, this term depends on temperature and enters the Hamiltonian, see page 427. This implies that temperature can enter into the classical mechanic's Hamiltonian. But the case is similar in quantum mechanics in that the Hamiltonian can also depend on the temperature. The details can be found in chapter 23, about page 452 where the harmonic Hamiltonian (which is part of the total Hamiltonian, and hence Schrödinger's equation) is given and depends implicitely on temperature. One way I see to justify this is that its eigenvalues (given on the same page) depend on temperature.

Maye an intuitive way to realize that the Schrödinger's equation depends on temperature in this case is by convincing oneself that the number of phonons is temperature dependent (it is shown in the book). The phonons can be thought as perturbations in the position of the ions, and these perturbations will affect electrons and the general Hamiltonian, hence Schrödinger equation.

If the example is too hard to grasp, consider the very simple following example. Our new system is now a crystal (say a metallic conductor) and we control its temperature as we wish. Everytime the temperature is below $100K$, we turn on an electromagnetic field. This field will have an influence on the electrons in the conductor and will have corresponding terms in the Hamiltonian (Schrödinger equation) as well. That potential certainly has a dependence on temperature, where the temperature is the one of the studied system. So this confirms that yes, it is quite possible to find systems in which Schrödinger equation involves temperature.

Nevertheless, it is not always possible to introduce temperature into Schrödinger equation. Like others have stated, if the number of particles of the system is too low or if the system is far from thermodynamics equilibrium, the concept of temperature is not defined, in which case it doesn't even make sense to speak of a temperature of the system. It's a different story if the number of particles is high enough and the system is only "slightly out of equilibrium" in which case it may prove useful to define temperature locally, in subregions of the whole system. This has been done by Onsager in 1931 and lead to the reciprocal relations that bear his name but this is going a bit off topic.

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  • $\begingroup$ Any bibliography of this? Do you think has any sense to define a potential depending on T? $\endgroup$ – user153036 Feb 26 '18 at 22:24
  • $\begingroup$ I've edited my answer to address your two queries. $\endgroup$ – thermomagnetic condensed boson Feb 27 '18 at 16:58
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It is quite late to answer this question, but I just stumbled about it, too, so I try to give an answer here.

I also think that it helps to consider firstly classical mechanics. Newton's equation of motion describes the motion of particles at fixed total energy (isolated system, NVE ensemble and 1st law of thermodynamics). If we consider describing let's say 1000 water molecules with Newton's equation and we initialize the system with zero velocities of all the water, the water will convert its potential energy to kinetic energy and some dynamic conversion of these two sources of energy will be started. Transfer of kinetic energy between individual water molecules should always happen by the principle of the 2nd law of thermodynamics, so the entropy of the complete system becomes larger, so that hot particles heat up cool particles. So the temperature is already in our simulations in the form of kinetic energy and we can also define it from that.

Now although we can define it, we cannot directly alter it. The reason for that is that our system is isolated, so there is no way to influence our water molecules from outside! That is the reason why the temperature does not appear as a variable in the Newton equation and also not in the Schrödinger equation, because both equation describe the motion of an isolated system. Inside this system there is temperature fluctuations, as said before, but we cannot change it from outside.

If we would want to change the temperature from outside, we need to include a heat bath/thermostat which heats up and cools down out water molecules by (fictitious) collisions.

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Schrödinger equation, just like Newton's laws, describes behavior of a single particle. Temperature is an average energy per particle, which is *meaningful only for large collections of particles.

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This questions is actually quite intriguing and I like to take a different perspective than the other authors, to maybe shine some light on this problem. Therefore, I consider time-dependent problems.

Let's start with the time-dependent Schrödinger equation (TDSE), with Hamiltonian $\hat{H}$ and many-body wave function $|\Psi(t)\rangle$, where the former is the generator of the dynamics and the TDSE reads

$$\text{i}\hbar\partial_t |\Psi(t)\rangle = \hat{H}|\Psi(t)\rangle, \hspace{0.4cm}|\Psi(t_0)\rangle=|\Psi\rangle$$

with some initial condition $|\Psi(t_0)\rangle$. The latter equation describe the real-time evolution of a quantum many-body system. The natural generalization of the latter to finite temperature is the Liouville-von Neumann (LvN) equation, which turns out be given as

$$\partial_t\hat{\rho}(t)=-\frac{\text{i}}{\hbar}[\hat{H},\hat{\rho}]$$

where $\hat{\rho}(t)$ is the density operator. At zero Kelvin, the LvN-equation is equivalent to the TDSE and the real-time dynamics of $\hat{\rho}$ are equivalently described by $|\Psi(t)\rangle$. However, at finite temperature, the density operator for a system in thermal equilibrium is related to a classical stochastic distribution over quantum mechanical states. The latter can be written in an energy eigenbasis $\{|n\rangle\}$ of the underlying Hilbert space as

$$\hat{\rho}_{th}=\frac{1}{Z_\beta}\sum_n e^{-\beta\varepsilon_n}|n\rangle \langle n|$$

In fact, the Hamiltonian entering the LvN eq. is identical to the Hamiltonian entering the TDSE, however, the nature of the corresponding state is fundamentally different. The density operator is the proper representation of a quantum mechanical system at finite temperature and not a wave function. Temperature does not enter the Hamiltonian at all. Moreover, quantum mechanical expectation values generalize at finite temperature to thermal ensemble averages, where the mean energy is for example given as

$$\langle\hat{H}\rangle=\text{tr}\{\hat{\rho}_{th}\hat{H}\}$$

where "tr" denotes the trace operation. The latter generalizes to time-dependent situations, when $\hat{\rho}(t)$ explicitly time dependent and $\hat{\rho(t_0)}=\hat{\rho}_{th}$, for example. Another important point relates to the time-evolution of the density operator. When the formal solution of the LvN-eq. is considered, it turns out, that ket states are propagated forward in time and bra states propagate backwards in time. This feature in combination with a doubled set of degrees of freedom is present in both quantum statistical dynamics as discussed here and finite temperature quantum field theory as given by Keldysh-Schwinger formalism, real-time path integrals and thermofield dynamics. In contrast, the TDSE leads only to a forward propagation of the wave function, while its adjoint is related to the backward evolution of a quantum mechanical system.

I now pick up environmental aspect mentioned by DanielSank. In fact, such a system-bath problem can actually be investigated from the perspective of the TDSE, when staying at $T=0K$, which is, however, numerically demanding. At finite temperature, this is again not possible due to the stochastic nature of the density operator, whereas temperature does not enter the Hamiltonian at all. Moreover, in those situations, one regularly augments the LvN-equation by an additional term, which takes into account the interaction with the surrondings and their temperature implicitly. This term, called dissipator, leads to an exchange of (thermal)energy from the system with the environment and renders the system dynamics non-unitary. I like to mention, that there exist ways (Liouville space formalism, thermofield dynamics) to express the LvN-equation in a way, that is formally equivalent to the TDSE, however, the basic properties do not change. Fields, where you can approach the microscopic relevance of temperature are from my perspective molecular junctions, heat and charge transport at the nanoscale or especially quantum thermodynamics and open quantum system theory.

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