# Energy is quantized

How can energy be quantized if we can have energy be measured like in 1.56364, 5.7535, 6423.654 kilo joules, with decimals? Thanks

Also isnt it quantization means energy is represented in bit quantities meaning you can not divide, lets say 1 bit of energy

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## 3 Answers

As far as the first part of your question goes, just having decimals in the number does not mean the energy levels are no longer quantized. Quantization of energy simply means that there are only specific energies that particles can take under certain circumstances. For example, you could say particle A can only have one of the following energies: {1.56364, 5.7535, 6423.654} kJ. Limiting the particle to these three energies is what it is meant by quantization of energy.

Also, there is no smallest bit of energy, for example the kinetic energy of a free particle can take a continuous range.

Mathematically, I am not certain how this is formulated. Off the top of my head, I would wager that any countable set could be considered quantized, but that would include all rationals which are dense in reals, so it really wouldn't be much of a quantization.

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tl;dr of how it's formulated: energies are eigenvalues of the Hamiltonian operator. The eigenvalues can be discrete (which is the technical term for being limited to selected values) or continuous. Jerry's answer has some more details. – David Zaslavsky♦ Sep 27 '12 at 4:37

Typically, in quantum mechanics, bound states are quantized and free/scattering states are not. This is because bound states, by the mere fact that they're constrained to a certain area, will have to satisfy certain boundary conditions, and these conditions won't be able to be satisfied in a continuous range.

The classic example of this is the infinite square well potential, where $V(x) = 0$ if $0<x<a$, and $V(x) =\infty$ elsewhere. Then, the particle will have zero probability of appearing outside of the well, and will have to satisfy the zero-potential Schrödinger equation $E\psi = -\frac{\hbar^{2}}{2m}\nabla^{2}\psi$ inside of the well. For simplicity, we'll only consider one-dimensional motion.

In this case, we see right away that the basis states to our solutions have to satisfy $\psi = A\sin\left(\frac{\sqrt{2mE}}{\hbar}x+\phi\right)$, and we also know that the wave function must be continuous, and that it is restricted to be zero for $x<0$ and $x>a$. We can satisfy the first boundary condition by choosing $\phi=0$, but the second one is not satisfied for all values of the energy. Instead, it is necessary that $\frac{\sqrt{2mE}}{\hbar}a=n\pi$, where $n$ is some integer. Thus, the allowed energies of 'pure' states of this system are quantized, and take the values $E_{n} = \frac{n^{2}\pi^{2}\hbar}{2m}$.

For any other bound state, you will find yourself using similar logic about boundary conditions, albiet with much, much more complexity. Note that, however, it is also the case that we can construct a general state out of the energy eigenstates $\Psi = \sum a_{n}\psi_{n}$, and that the expectation value for the energy of $\Psi$ will be $\sum|a_{n}|^{2}E_{n}$, so the values for the "average" value of a state are still allowed to be continuous (and in the case of the infinite square well, can actually take any value greater than the ground state energy).

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I see that the wave function must be continuous as a postulate, but is there an explanation for why that must be true? – jcohen79 Sep 27 '12 at 4:22

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The momentum operator is $-i\hbar\partial/\partial x$. If the wavefunction is spatially discontinuous that would imply infinite momentum. – John Rennie Sep 27 '12 at 9:27

An excellent way to understand how the wave function works in quantum mechanics is to study the model of the hydrogen atom. We can see in this model that the quantum variables $n,l,m$ are effectively variables that determine the shape of the spatial density associated with the detection of an electron. The quantum aspect is that the variables $n,l,m$ are integers, where the continuous aspect is density associated with the wave function. It is important to understand that the density function is a space filling function. This means that there is a value of the function associated with each point in space.

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