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The product of annihilation operator with zero ground state of a harmonic oscillator is zero since energy cannot be negative or less than zero. I understand this explanation but what is the intuitive explanation of its conjugate operator i.e creation operator when operating on bra is also zero ?

I understand this part, as the annihilation operator decreases energy levels and zero is our last energy level, we write it's a product as zero. i.e $$\hat{a}|0\rangle=0$$

But how can I explain this? i.e the creation operator acting on the bra is also zero.ie.

$$\langle 0|\hat{a}^\dagger=0$$

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  • $\begingroup$ Can you explain your notation a little bit? $\endgroup$ – Young Kindaichi Oct 27 '20 at 11:21
  • $\begingroup$ How good is your understanding of Dirac Notation, Bra's, Ket's and adjoints of operators ? It seems that your confusion stems from a lack of that, and a proper understanding of these topics should easily answer your question. $\endgroup$ – Hans Wurst Oct 27 '20 at 11:38
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The statement that $a|0\rangle=0$ is exactly the same as the statement $\langle 0|a^\dagger=0$. This is one of these things that is easier to see in the mathematicians notation. In their notation, the inner product between two vectors $u$ and $v$ is denoted by $\langle u,v\rangle$. The adjoint of an operator $A$ is defined by $\langle u, Av\rangle=\langle A^\dagger u,v\rangle$. Now, let us denote the vector $|0\rangle$ by $u$. Then for every vector $v$, which in ket-bra notation we may denote by $|v\rangle$, we have $$\langle 0|a^\dagger|v\rangle=\langle u, a^\dagger v\rangle=\langle au,v\rangle=\langle 0,v\rangle=0.$$ The first equality is just a change of notation. The second is the definition of adjoint of an operator. The third is the fact that the annihilation operator annihilates the vacuum, i.e. $a|0\rangle=0$, or, in our notation, that $au=0$. Here a word of caution: when going between the mathematicians and the physicists language one must be careful not to confuse the vacuum $|0\rangle$ with the $0$ of the Hilbert space. This is why we had to introduce the new label $u$. Finally, the fourth equality is a standard property of inner products. It can be easily derived from the skew linearity $\langle k u,v\rangle=\overline{k}\langle u,v\rangle$ for all vectors $u$ and $v$ and scalars $k$. In particular, take $k=0$.

We conclude that for all $|v\rangle$ we have $\langle 0|a^\dagger|v\rangle$. Therefore, the operator $\langle 0|a^\dagger=0$. Indeed, an operator is equal to $0$ if and only if its action on all of the vectors vanishes.

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I'm not aware of any "intuitive" physical reason why this must be, whatever that may mean. This is a straight mathematical application of the standard results ($\hat A$ and $\hat B$ are operators): $$(AB)^\dagger=\hat B^\dagger \hat A^\dagger \tag{1},$$ and $$(\langle n|m\rangle)^\dagger=\langle m|n\rangle. \tag{2}$$ This then implies: $$\hat a|0\rangle=0\rightarrow\left(\hat a|0\rangle\right)^\dagger =(|0\rangle^\dagger)(\hat a)^\dagger=\langle0|\hat a^\dagger=0. \tag{3}$$ Note that this is required to equal zero since $ 0^\dagger= 0$, where, in slightly poor (but standard) notation we implicitly mean "the zero vector" when we write $0$.

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  • $\begingroup$ Usually one uses $(\langle n\vert m \rangle)^*=\langle m\vert n\rangle$ rather than the $^\dagger$ version you have. The “transpose” part is implied by trading the bra with the ket. $\endgroup$ – ZeroTheHero Oct 27 '20 at 12:25
  • $\begingroup$ @ZeroTheHero $\langle n |m \rangle$ is a number so $\dagger$ and $*$ coincide. $\endgroup$ – jacob1729 Oct 27 '20 at 13:27

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