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Apparently bound states in quantum mechanics require energy states to be discrete. That means energy in such systems is quantized, right? However, say that we have a superposition of energy eigenstates. Using the superposition principle, I say that their average is a possible state, since we also say that there's the same probability that we'll observe each of the two available states. Then since this average is allowed, I can take the average of that and the smaller of the original states, can't I? And if I repeat this an infinite number of times, I should get something which is as good as continuous, right? If I talk about the $n$th energy levels of some system, I won't be able to express the energy each of these allowed states in terms $n$ without making some messy equations. Isn't this a contradiction?

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    $\begingroup$ Possible duplicate of What is the energy of a superposition of energy eigenstates? Not exactly the same, but the discussion of expectation values should resolve this question. $\endgroup$ – Chair Feb 8 at 15:07
  • $\begingroup$ On second thoughts, I don't think this is a duplicate. There is something to discuss in the difference between expectation values and superposition states. Voting to reopen. $\endgroup$ – Chair Feb 9 at 13:40
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This can be understood from a purely classical perspective, and it's a notion which appears in the discussion of almost any discrete variable. The expectation value is not necessarily an allowed state. For instance, even with a fair die, the expectation value is 3.5 (since it's the average of the possible numbers), but you'll obviously never get 3.5.

So imagine you're dealing with a superposition of two energy eigenstates $|n_1\rangle$ and $|n_2\rangle$. The question describes a long process of averaging the energies of those states and then using that average with either $|n_1\rangle$ or $|n_2\rangle$, but that isn't even necessary to get an expectation value of some arbitrary value in between the two 'allowed' states. In your discussion of 'averages', I assume you're talking about something like this: $$\frac{1}{\sqrt{2}}|n_1\rangle+\frac{1}{\sqrt{2}}|n_2\rangle$$ But you don't need to do that whole infinity-many-averages thing. You could simply consider $$\alpha |n_1\rangle+\beta n_2\rangle$$ where $$|\alpha|^2 +|\beta|^2=1$$ and choose $\alpha$ and $\beta$ such that the expectation value is whatever you want between $|n_1\rangle$ and $|n_2\rangle$.

However, in all of this, whatever $\alpha$ and $\beta$ you consider, you will only ever observe either $|n_1\rangle$ or $|n_2\rangle$. Nothing in between.

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  • $\begingroup$ Why did you take square roots of 1/2 in the superposition of $|n\rangle$ states though? $\endgroup$ – JumpingChimp Feb 11 at 13:09
  • $\begingroup$ The sum of the coefficients' squares must be 1. Note the other equation $|\alpha|^2 +|\beta|^2=1$ $\endgroup$ – Chair Feb 11 at 13:12
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No. It is a good thought but no. In short, you are correct that we can prepare states whose average energy takes on any value between the lowest and highest energy state. But this is different than creating a state which is "between" the energy eigenstates.

Let $|n\rangle$ be the $n$th energy eigenstate. This means that we have that:

$$\hat{H}|n\rangle = E_n |n\rangle$$

and better yet:

$$E_{avg} = \langle \hat{H}\rangle = \langle n|\hat{H}|n\rangle = E_n$$

We can define a state:

$$ |\psi\rangle = a|0\rangle + b|100\rangle $$

with $|a|^2 + |b|^2 = 1$

$$ E_{avg} = \langle \hat{H}\rangle = \langle \psi | \hat{H}|\psi\rangle = |a|^20 + |b|^2 100 = 100 |b|^2 $$

Since $0<|b|<1$ we see that we can have a state which has any average energy between $0$ and $100$. In fact, since $|b|$ is complex there are already an infinity of states which have this property.

Suppose $b = \frac{1}{\sqrt{2}}$ so that $|b|^2 = \frac{1}{2}$ and $E_{avg} = 50$. This means that if you prepare and measure this state many times that half the time you will measure $0$ and half the time you will measure $100$. Contrast this with the state $|50\rangle$ where you would measure $50$ every time.

This is meant to illustrate the difference between a quantum state whose average energy is $50$ and an eigenstate of energy whose measured energy is always 50.

There are other differences as well. Suppose the system is a harmonic oscillator. In that case $n$ of state $|n\rangle$ tells you that the oscillator has an amplitude of oscillation of exactly $\sqrt{n}$ in phase space ($\hat{X}^2+\hat{P}^2 = \hat{n} \rightarrow n$ in the appropriate units). If the system is in state $|\psi\rangle$ above you will find it half the time with amplitude $0$ and half the time with amplitude $10 = \sqrt{100}$ but never with amplitude $\sqrt{50}$. This is another sense in which the superposition of two states is not really like the average of the two states. In terms of expectation values things can sometimes be similar but not always. Generally what you've described is not a good way of thinking about quantum states.

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  • $\begingroup$ When you mention the state $|100\rangle$, are you talking about some random state with the energy of 100 (units)? Or is it the name of the state? If it's actually for an energy of $100$, then is $|0\rangle$ supposed to be zero-energy? $\endgroup$ – JumpingChimp Feb 11 at 13:05
  • $\begingroup$ Yes, $|100\rangle$ and $|0\rangle$ are energy eigenstates with $100$ and $0$ Energy respectively. (Ignoring zero point energy in the case of a harmonic oscillator which contributes energy of 1/2 unit) $\endgroup$ – jgerber Feb 11 at 15:35
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As the Schrödinger equation is linear, any linear combination of solutions is also a solution. However, solutions have be normalised to unity, thus representing unit elementary charge. Furthermore, most of these solutions are not stationary. Only solutions of the time independent equation are and represent discrete energies. Non stationary solitions can represent energies intermediate between those of the stationary states.

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  • $\begingroup$ So I can have a wavefunciton evolve over time and eventually I must be able to observe all energies at some point of time or other? $\endgroup$ – JumpingChimp Feb 11 at 13:08
  • $\begingroup$ No you will observe these only on average. A single measurement outcome can only be one of the eigenvalues. $\endgroup$ – my2cts Feb 11 at 15:17

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