 # Thanks to Zarah.

## Presentation on theme: "Thanks to Zarah."— Presentation transcript:

Thanks to Zarah

Getting to Schrödinger’s wave equation
Works for light (photons), why doesn’t it work for electrons? We found that solutions to this equation are or with the constraint which can be written Multiplying by ħ we get which is just E=pc But we know that E=pc only works for massless particles so this equation can’t work for electrons.

Getting to Schrödinger’s wave equation
doesn’t work for electrons. What does? or Note that each derivative of x gives us a k (momentum) while each derivative of t gives us an w (energy). Equal numbers of derivatives result in E=pc For massive particles we need kinetic energy So we need two derivatives of x for p2 but only one derivative of t for K. If we add in potential energy as well we get the Schrödinger equation…

= + The Schrödinger equation Kinetic energy Potential energy
The Schrodinger equation for a matter wave in one dimension (x,t): Kinetic energy Potential energy Total energy + = This is the time dependent Schrödinger equation (TDSE) (discussed in 7.11) and is also the most general form. This potential energy is a function of x and t. It gives the potential energy of the particle for any x and t. It is not intrinsic to the particles but something from the problem at hand.

Time independent Schrödinger equation
In most physics situations (like hydrogen atom) the potential function U does not change in time so can write U(x,t) = U(x). In this case, we can separate Y(x,t) into y(x)f(t): We will then use the time independent Schrödinger equation (TISE) for the x-component of the wavefunction (lower case psi):

Given a potential energy function U(x), where would a particle naturally want to be?
Where U(x) is highest Where U(x) is lowest Where U(x) < kinetic energy Where U(x) > kinetic energy Does not depend on V(x) Particles want to go to position of lowest potential energy, like a ball going downhill. U(x) x

The infinite square well (particle in a box)
The potential energy function is x < 0: U(x) ≈ ∞ x > a: U(x) ≈ ∞ 0 < x < a: U(x) = 0 Potential Energy x a This is called the infinite square well (referring to the potential energy graph) or particle in a box (since the particle is trapped inside a 1D box of length a. We are interested in the region 0 < x < a where U(x) = 0 so Becomes (for the states in the box)

Guess a solution to How about y(x) = Acos(kx)? which gives or The total energy E is completely kinetic energy (because we set the potential energy U=0)

Infinite square well solution
The functional form of the solution is a V(x) Now we apply the boundary conditions or LHS: x = 0: 1 2 We also know so that

For an infinite square well, what are the possible values for E?
a V(x) Any value (E is not quantized) Putting into the TISE gives so (just kinetic energy) Putting in the k quantization condition gives

Infinite square well (particle in a box) solution
After applying boundary conditions we found and which gives us an energy of What is the lowest energy possible? Something else

Infinite square well (particle in a box) solution
After applying boundary conditions we found and which gives us an energy of x U=0 a E1 4E1 9E1 16E1 n=4 n=3 n=2 n=1 Energy Things to notice: Energies are quantized. Minimum energy E1 is not zero. This is a general principle of QM.

Infinite square well (particle in a box) solution
After applying boundary conditions we found and which gives us an energy of Things to notice: x V=0 a E1 4E1 9E1 16E1 n=4 n=3 n=2 n=1 Energy Energies are quantized. Minimum energy E1 is not zero. Consistent with uncertainty principle. x is between 0 and a so Dx~a/2. Since DxDp≥ħ/2, must be uncertainty in p. But if E=0 then p=0 so Dp=0, violating the uncertainty principle. When a is large, energy levels get closer so energy becomes more like continuum (like classical result).

A grain of sand Suzy gently places a tiny grain of sand at the bottom of a very narrow and deep well. She says: “Because of the laws of QM, this grain of sand have a finite energy so it must be floating off the ground”. Liz says: “That is ridiculous – of course it is not levitating off the ground. Therefore sand must not be quantum.”. Who is correct? x U=0 a E1 4E1 9E1 16E1 n=4 n=3 n=2 n=1 Energy Suzy Liz Neither – sand is “quantum” but “finite energy” does not mean that the sand is levitating, which would mean U is larger.