3
$\begingroup$

I'm studying quantum mechanics in its most basic level (I don't even know if Physicists call this already quantum mechanics) and I have one doubt in Schrodinger's equation. The book presents the equation for the special case where the solution is of the form $\Psi(x,t)=\psi(x)e^{-i\omega t}$ and says that Schrodinger's equation (in one dimension) is

$$\dfrac{d^2\psi}{dx^2}+\dfrac{8\pi^2 m}{h^2}(E-U(x))\psi(x)=0$$

Where $m$ is the mass of the particle, $E$ it's total energy and $U$ it's potential energy function.

The first doubt that arises is the following: the book says that $E$ is the "constant total energy" of the particle. But wait, since $E = K + U$ and $U$ varies, clearly $E$ should vary. How can $E$ be constant if $U$ is not?

Also, when we write $E-U(x)$ isn't this simply $K$, the kinectic energy? Why do we bother then writing explicitly $E-U$?

I feel that the potential $U(x)$ on the equation and the one that is part of $E$ are different, but I'm not understanding how.

$\endgroup$
  • 2
    $\begingroup$ If U varies, U being the work done against the forces to exist in x, then energy has been lost or gained of the opposite amount by a particle to exist in x, in the form of kinetic energy. This states the kinetic energy theorem. $$\Delta K = - \Delta U $$ by definition of the physical meaning of potential energy. Hence $$\Delta E = 0$$. Here $E = \hbar \omega $$ is a constant as the interaction is time independent. $\endgroup$ – JumpArtist May 21 '14 at 0:13
  • 2
    $\begingroup$ I forgot to reply "yes" to your question in the title. Written so, your equation is an eigenvalue problem for the kinetic energy operator if you want to see it like that. E-U(x) being the kinetic energy eigenvalue in the position representation. $\endgroup$ – JumpArtist May 21 '14 at 0:21
4
$\begingroup$

Note that the equation you give can be trivially restated as $$\left[-\frac{\hbar^2}{2m}\frac{\mathrm{d^2}}{\mathrm{d}x^2}\right]\psi(x) = \left[E-U(x)\right]\psi(x)\text{,}$$ where $\hbar = h/2\pi$. So if in this situation we interpret kinetic energy as the difference between total and potential energy, this tells us that the kinetic energy operator must in the position basis be: $$\hat{T} = -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\text{.}$$ We can also definite a potential energy operator, which in this basis is just trivial multiplication: $\hat{U}\left[\psi(x)\right] = U(x)\psi(x)$.

The first doubt that arises is the following: the book says that $E$ is the "constant total energy" of the particle. But wait, since $E = K + U$ and $U$ varies, clearly $E$ should vary. How can $E$ be constant if $U$ is not?

What the books means by "constant total energy" is that the total energy has a definite value, i.e., the given wavefunction is an eigenstate of the total energy: $$\left[\hat{T} + \hat{U}\right]\psi = E\psi\text{.}$$ If one measures the energy of the particle correctly, one will get this definite value $E$. This won't happen for either the kinetic or potential energies, except in some trivial cases: measuring them will give a random eigenvalue of the corresponding operators.

However, given an energy eigenstate as we have here, the kinetic and potential energies are time-independent in the sense that the probability distribution of the possible measurement results does not vary in time. The same holds for any observable that's not explicitly dependent on time.

Also, when we write $E-U(x)$ isn't this simply $K$, the kinectic energy? Why do we bother then writing explicitly $E-U$?

That wouldn't make sense. If we simply write $K$ as a constant in the equation you give, that would suggest that the kinetic energy takes on a definite value, and this is not true for the wavefunction given. To have a definite value of kinetic energy, the state $\psi$ must be an eigenstate of kinetic energy, $\hat{T}\psi = K\psi$.

$\endgroup$
4
$\begingroup$

You are correct in searching the energy balance connection between the classical and the quantum formulation but writing the equation in a manner that does not help. This is better:

shcroedinger equation

The kinetic and potential energies are transformed into the Hamiltonian which acts upon the wavefunction to generate the evolution of the wavefunction in time and space. The Schrodinger equation gives the quantized energies of the system and gives the form of the wavefunction so that other properties may be calculated.

Remember , in quantum mechanics the operators operate on the wave function to give an eigenvalue, a number that can be measured, for the operator's function. In this case, the total energy operator is the Hamiltonian, which is composed by the kinetic and potential energy operators.

$\endgroup$
1
$\begingroup$

When dealing with such Schroedinger equation, one is interested in finding the eigenvalue $E$ and eigenfunction $\psi$. The relation $E = K+U$ does not appear here. The equation contains expression $E-U(x)$, which is a function of configuration $x$, but nothing about kinetic energy is meant. $E$ is just eigenvalue of the Hamiltonian and $U(x)$ is potential energy function.

To give $E$ interpretation is difficult. If we say it is energy of the particle, then in places $x$ where the number $E-U(x)$ is negative the relation $E = K+ U(x)$ from classical mechanics does not hold, since kinetic energy $K$ must be positive. In quantum theory people talk about total energy $E$, but rarely about kinetic energy.

The other possibility is to assume that the relation $A = K+U(x)$ is valid where $A,K,U$ are actual total, kinetic, potential energy, and the number $E$ in the Schroedinger equation is not instantaneous energy of the particle, but only is somehow connected to it, say, it is expected average value of energy.

$\endgroup$
0
$\begingroup$

"Isn't E−U=K in Schrodinger's Equation?"

Yes, it is.

"Why do we bother then writing explicitly E−U?"

From a practical point of view, we do not know $E$ and $K$. However, from calssical physics we have a lot of scenarios where U is analyticaly known. So if we want to investigate a problem on a quantum mechanics level, we put in the $U$ that corresponds to the problem and solve it so find $E$ ($K$ is rarely a thing of interest). It is much more difficult to write down $K$ for a given problem that isn't already solved, e.g. what is K for a single electron surrounded by atoms? The $U$ for this situation is written down easily.

But wait, since E=K+U and U varies, clearly E should vary. How can E be constant if U is not?

You are mistaken here. The equation you present assumes that the Hamiltonoperator isn't timedependent, which already means that $E$ is constant for all times. This is "achieved" by a varying $K$. $K(x)$ and $U(x)$ vary in such a way that their sum (also called $E$) has always the same value. Imagine an electron that is moving through a Potential $U$ but with varying speed and therefore varying $K$.

$\endgroup$
  • $\begingroup$ You should read Stan Liou's answer to see why your last two paragraphs are wrong. $\endgroup$ – Kyle Kanos May 21 '14 at 13:10
  • $\begingroup$ could you plz be more explicit here, I dont see where anything he writes would contradict with my answer $\endgroup$ – LeFitz May 21 '14 at 18:19
  • $\begingroup$ $K$ is the operator (Stan defines it as $\hat{T}$, but it's the same thing); this is known contrary to your assertion. It's also a "thing of interest" since it's how we get the energy eigenvalues. $\endgroup$ – Kyle Kanos May 21 '14 at 18:23
  • $\begingroup$ I believe you misunderstood my answer. I didnt say we dont know the $\hat{T}$ operator, I meant we dont know his spectrum $K$ and we are rarely interested in it. We are interested in $\hat{H}$ and his spectrum $E$ which is simply why the eigenequation $\hat{T}\psi = K\psi$ is seldom used and therefore there is no real need to relable $E-U$ as $K$ as intended by OPs question. Also I didnt say it is NO thing of interest, I said it is RARELY a thing of interest. $\endgroup$ – LeFitz May 21 '14 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.