Mathematical basics of quantum mechanics

I study QM, in particular I need to implemend the computation of vibrational spectrum for polyatomic molecules by forca constants in the "rigid rotor - harmonic oscillator" approximation in my program Chemcraft. I have watched some videos from the channel "Professor Dave explains".
The time-independent Schrodinger equatiuon is as follows:

$$-\frac{\hbar^2}{2m}\frac{d^2(\psi (x))}{dx^2}+U \psi (x)=E \psi (x)$$

Firstly I tried to solve the SE for a particle in a potential box. We have $$U(x)=\infty$$ for x<0 and x>a, and U(x)=0 for x from 0 to a. So:
$$\frac{d^2 \psi(x)}{dx^2}=-k^2 \psi(x)$$
$$k=\frac{\sqrt{2mE}}{\hbar}$$

We have a function, which turnes to itself but with negative sign after double differentiation. As far as I understand, only sinus and cosine have this feature, and also an exponent from x*i.
Because $$\psi(0)=0$$, we can reject cosine; because $$\psi(a)=0$$, we can write after normalization ($$\int_0^a(\psi^2(x)dx)=1$$):

$$\psi_n=\sqrt{\frac{2}{a}}\sin(\frac{n \pi}{a}x)$$

I have two quesions:
1) Can we write the superposition of the functions above as another correct solution for the SE?
$$\psi=A\sin(\frac{\pi}{a}x)+B\sin(\frac{2 \pi}{a}x)+C\sin(\frac{3 \pi}{a}x)$$
2) Can we prove that the correct solution does not include imaginary numbers?

1) Yes.
2) No, because A, B, C, ... in 1) can be imaginary.

Spathi said:

I have two quesions:
1) Can we write the superposition of the functions above as another correct solution for the SE?
$$\psi=A\sin(\frac{\pi}{a}x)+B\sin(\frac{2 \pi}{a}x)+C\sin(\frac{3 \pi}{a}x)$$

Demystifier said:

1) Yes.

No, because it is no longer an Eigenfunction. Each Eigenfunction has a unique energy ##E##, but if you sum functions with different ##E##, then the resulting function no longer has an ##E## with which Schrödinger's equation would be satisfied.

Demystifier said:

1) Yes.

gentzen said:

No, because it is no longer an Eigenfunction.

Is one of you talking about the TDSE and the other the TISE?

Nugatory said:

Is one of you talking about the TDSE and the other the TISE?

Well, maybe a better answer is that the superposition is not a solution of the TISE, and that you have to include the time dependency before you can use the superposition as a solution of the TDSE:
$$\psi=A\sin(\frac{\pi}{a}x)\exp(\frac{-i\hbar}{2m}\frac{\pi^2}{a^2}t)+B\sin(\frac{2 \pi}{a}x)\exp(\frac{-i\hbar}{2m}\frac{4\pi^2}{a^2}t)+C\sin(\frac{3 \pi}{a}x)\exp(\frac{-i\hbar}{2m}\frac{9\pi^2}{a^2}t)$$

Demystifier said:

No, because A, B, C, ... in 1) can be imaginary.

This is correct for TDSE but not for TISE?

Spathi said:

2) Can we prove that the correct solution does not include imaginary numbers?

If you are studying QM, it's best to lose any squeamishness about complex numbers.

Nugatory said:

Is one of you talking about the TDSE and the other the TISE?

Obviously.

Spathi said:

Can we prove that the correct solution does not include imaginary numbers?

Demystifier said:

No, because A, B, C, ... in 1) can be imaginary.

Spathi said:

This is correct for TDSE but not for TISE?

Because there were no complex numbers in your specific TISE, what you can do is to take any solution, and take its real or imaginary part to get a solution not including imaginary numbers.
And because you can get back that complex solution as superposition of those two real solutions, it is clear that looking at real solutions is enough for your specific TISE.

I continue viewing the videos on the channel "Professor Dave explains", in particular the video about quantum tunneling:



Let's consider we have a particle without any forces (U=0).
$$\frac{d^2 \psi(x)}{dx^2}=-k^2 \psi(x)$$
$$\psi (x)=A e^{-i k x}+B e^{i k x}$$
And then Dave says: because we know that the particle moves from the left to right, not vice versa, we can eliminate the B term:
$$\psi (x)=A e^{-i k x}$$
I want to understand how this works. Where in this formula is the postulation that in the past the particle was at the left, in the future it will be at the right?

I understand that we can write the TDSE as follows:
$$\psi (x,t)=C e^{-i k x-iwt}$$
If we take time interval dt>0, the whole function will shift to right. But if the dt=2Pi/w, nothing will change. Can you help me understand all this?

Spathi said:

I want to understand how this works. Where in this formula is the postulation that in the past the particle was at the left, in the future it will be at the right?

The equation you start with includes all possible solutions. If you fire an electron from the left, then that is a more specific physical scenario. It's essentially an initial condition of your experiment.

You could equally fire the electron in from the right.

Obviously they calculate scattering states in 1D potential-well problems, which is a very good exercise by the way.

Let's discuss the potential in the OP's figure with a little simplification, i.e., making the square well symmetric around ##x=0##, i.e.,
$$V(x)=\begin{cases} V_0 & \text{for} \quad |x|<a/2,\\ 0 & \text{for} \quad |x|>a/2.\end{cases}$$
Then the whole problem is parity invariant, i.e., spatial reflections ##x \rightarrow -x## is a symmetry.

Then there is a degeneracy, i.e., for each energy level ##E>0## there is a solution with positive and a solution with negative parity, i.e., ##u_E^{pm}(x)## with ##u_E^{(\pm)}(-x)=\pm u_E^{(\pm)}(x)##. These common eigensolutions of
$$\hat{H}=\hat{p^2}{2m} + V(x)$$
and the parity operator
$$\hat{P}\psi(x)=\psi(-x),$$
are, however not what you want in scattering problems. There you usually have in mind a particle coming from one side, hitting the potential and you want to know the probability for finding the particle on the other side of the potential. Then you need a solution, which has an "outgoing" wave on the right only, while on the left side you have a superposition of the incoming and the reflected wave. The corresponding amplitudes for the reflected and emitted wave (modulus squared) are the probabilities for reflection and transmission.

Due to the parity symmetry, it's of course completely equivalent to let the particles come from the right and going through to the left (+being reflected to the right by the potential). This "left-moving" solution is just the parity-transformed state of the above described "right-moving" solution and thus gives the same reflection and transmission probabilities.

As I said, it's a good exercise to find all the solutions with these properties. It's a bit easier to look for the parity eigenstates first, because then you need to consider only half of the boundary conditions at ##x=a/2##, while for the left- and right-moving states you need to consider both boundary condtions at ##x=a/2## and ##x=-a/2##.

I still don' understand. If the potential is zero, we have

$$\psi (x)=A e^{-i k x}+B e^{i k x}$$

As far as I understand, the A term means that the particle moves from left to right, the B that it moves from right to left. For a particle in a box, as far as I understand, we have both terms; will that be correct to say that we have a superposition of particles moving in both directions?
Where is the speed of the particle in this formula?
I don't remember the Maxwell's equations; maybe they are better than the equaion above for describing the physical reality? I mean that the at least contain the speed of light constant.

It's the other way. The separation ansatz for the time-dependent wave function is
$$\Psi(t,x)=\exp(-\mathrm{i} \omega t) \psi(x).$$
The plane-wave solution ##\psi(x) \propto \exp(+\mathrm{i} k x)## means that the wave moves to the right (for positive ##k##) and ##\psi(x) \propto \exp(-\mathrm{i} k x)## means it moves to the left (for positive ##k##).

For a particle in an infinite square well there's no interpretation of ##k## in terms of momenta and there are no "moving" solutions.

For the finite potential step, we discuss here, you have a momentum operator and ##\pm k## for ##k>0## refer to particles moving to the right/left, respectively.

Spathi said:

For a particle in a box, as far as I understand, we have both terms; will that be correct to say that we have a superposition of particles moving in both directions?

Yes, precisely. Even in classical mechanics, that's what you have statistically at any point in time. QM takes that statistical uncertainty to a new level: fundamentally the particle is not at any precise point, nor has it a precise momentum unless you measure it. It's state is a probability distribution for these measurement results.

Spathi said:

Where is the speed of the particle in this formula?

The speed is related to the momentum.

Spathi said:

I don't remember the Maxwell's equations; maybe they are better than the equaion above for describing the physical reality? I mean that the at least contain the speed of light constant.

This makes no sense. Maxwell's equations relate to classical electromagnetism.

I am continuing watching Dave's videos, in particular about the barrier:



Here are his formulas:$$\psi (x)=A e^{i k_0 x}+B e^{-i k_0 x}$$

For x<0

$$\psi (x)=C e^{i k_0 x}$$

For x>a

$$\psi (x)=F e^{k_b x}+F e^{-k_b x}$$

For 0<x<a

$$k_0=\frac{\sqrt{2m E}}{\hbar}$$

$$k_b=\frac{\sqrt{2m (U_0 - E)}}{\hbar}$$

The boundary conditions (equality of the wavefunctions and their first derivatives) produce theese equations:

$$A+B=F+G$$
$$ik_0 A -ik_0 B=k_b F - k_b G$$
$$F e^{k_{b}a}+G e^{-k_{b}a}=C e^{i k_0 a}$$
$$k_b F e^{k_{b}a}-k_b G e^{-k_{b}a}=i k_0 C e^{i k_0 a}$$

The solution of this system leads to

$$\frac{C}{A}=(1+\frac{\sinh^2(k_b a)}{4 \eta(1-\eta)})^{-0.5}$$
$$\eta=\frac{E}{U_0}$$

Then Dave performs calculations with this formula, and gets the following: if the barrier height is 1 micro-electronvolt and the barried thickness is 0.1 nanometer, then the tonnel effect is clearly seen (e.g. if the electron energy is 20% of the barrier height, nearly 20% of electrons pass the barrier).
If the electron energy higher than the barrier, we get this formula:

$$\frac{C}{A}=(1+\frac{\sin^2(k_b a)}{4 \eta(1-\eta)})^{-0.5}$$

This formula shows the second interesting effect - oscillations (it includes a sin wave).
I am afraid that Dave was wrong with the calculations; anyway, are these two formulas correct?

I have found an error in Dave’s video:



Trying to calculate all myself by his formula:

$$\frac{C^2}{A^2}=(1+\frac{\sinh^2(k_b a)}{4 \eta(1-\eta)})^{-1}$$
$$\eta=\frac{E}{U_0}$$
$$k_b=\frac{\sqrt{2m (U_0 - E)}}{\hbar}=\frac{\sqrt{2m U_0 (1- \eta)}}{\hbar}$$We have the barrier height $$0.4 eV=6.40872 \cdot 10^{-20} J=6.40872 \cdot 10^{-20} kg \cdot m^2/c^2$$, barrier thickness $$1nm=10^{-9} m$$, electron mass $$9.109 \cdot 10^{-31} kg$$, $$\hbar=1.055 \cdot 10^{-34} J \cdot c=1.055 \cdot 10^{-34} kg \cdot m^2/c$$, $$\eta=0.5$$$$K_b=\frac{\sqrt{2 \cdot 9.109 \cdot 10^{-31} kg \cdot (6.40872 \cdot 0.5) \cdot 10^{-20}kg \cdot m^2/c^2}}{1.055 \cdot 10^{-34} kg \cdot m^2/c}=1,62009 \cdot 10^9 m^{-1}$$$$\frac{C^2}{A^2}=(1+\frac{sinh(1,62009 \cdot 10^9 m^{-1} \cdot 10^{-9}m)}{4 \cdot 0.5 \cdot (1-0.5)})^{-1}=0.73$$
So, I found that the tunnel effect is clearly seen if the barrier thickness is 1 nm and the barrier energy is 0.4 eV. Is that correct?

I am continuing watching Dave (harmonic oscillator):

This blog provides a relatively convenient format, although I am afraid the author can produce errors.

We have SE:
$$-\frac{\hbar}{2m}\frac{d^2 \psi(x)}{dx^2}+\frac{1}{2}k x^2 \psi(x)-E \psi(x)=0$$
For convenience we change x to X:
$$\frac{d^2 \psi(X)}{dx^2}+\varepsilon \psi(X)-X^2 \psi(X)=0$$$$X=x/ \alpha$$
Then Dave says that this equation is solved via Hermite polynomials:I don’t understand, where does epsilon go? Can you help me understand this?

Well, you can check that the wavefunctions satisfy the original equation ...

##\ ##

BvU said:

Well, you can check that the wavefunctions satisfy the original equation ...

Yes, and that will show you how ##n## and ##\epsilon## are related to each other.