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This is a well-known result of ladder operators, which obviously means that you can't remove energy from the vacuum. But what is $\hat{a}|0\rangle=0$ actually saying? How does it say the "you can't" part of the sentence? My best guess is "The probability of having such a state is $0^2=0$ ", but I'm not sure.

A satisfactory answer to this must contain the word "zero", since that's the only thing that is given to us by the formula. Sentences like "you can't do that" or "that doesn't exist" aren't good translations of what the formula is saying. The formula is saying that some quantity is zero, and my question is: What is that quantity?

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    $\begingroup$ Isn't it more of you can't destroy particles that aren't there? $\endgroup$ – Kyle Kanos Jan 23 '17 at 2:04
  • $\begingroup$ But how does it say the "you can't" part? What does the 0 mean? $\endgroup$ – Pato Raimundo Jan 23 '17 at 2:06
  • $\begingroup$ It says that with the fact that the operation results in 0. $\endgroup$ – Kyle Kanos Jan 23 '17 at 2:08
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    $\begingroup$ See my answer to physics.stackexchange.com/q/112807 $\endgroup$ – JamalS Jan 23 '17 at 2:09
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    $\begingroup$ Doesn't is say "There is no such state arrived at by destroying the vacuum state"? $\endgroup$ – garyp Jan 23 '17 at 2:28
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The formula says that the result of applying $a$ to the state $|0\rangle$ (which is a physical state) is $0$ (which is not a physical state). Since we know that $a$ is the destruction operator, what this equation implies is that there is no physical state which has less particles than $|0\rangle$; therefore, $|0\rangle$ is to be interpreted as a state with zero particles.

We say that "you can't take particles out of the vacuum" because if you attempt to calculate $a|0\rangle$, you get something which is not a physical state (for example, it's not normalizable) or, if you prefer, something that has zero overlap with every other vector.

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  • $\begingroup$ I liked the last thing you said, could you expand on it? What are the implications of a vector having zero overlap with all other vectors? $\endgroup$ – Pato Raimundo Jan 23 '17 at 4:13
  • $\begingroup$ @PatoRaimundo: For any state $|\psi\rangle$ at all, the inner product of $0$ (not $|0\rangle$!) and $|\psi\rangle$ is zero. This means that no matter the initial state of your system, the probability to ever find it in the state $0$ is, well, $0$, so that state is inaccessible. $\endgroup$ – Javier Jan 23 '17 at 12:56
  • $\begingroup$ Great! That's the answer I was looking for. The equation is saying: the state $\hat{a}|0\rangle=|?\rangle$ is a function or vector that is just zero everywhere. Therefore, for any state, the probability that they are the same is $\langle ?|\psi\rangle=\textbf{0}|\psi\rangle =0$, even with itself: $\langle?|?\rangle =0$. So it's saying "the probability of the system being is such a state is zero". If you could include something like that in your answer for more people to see, then I'll accept it. $\endgroup$ – Pato Raimundo Jan 24 '17 at 8:30

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