# Why product of creation operator with ground state enegy of vacuum is zero?

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.

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

• Can you explain your notation a little bit? – Young Kindaichi Oct 27 '20 at 11:21
• 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. – Hans Wurst Oct 27 '20 at 11:38

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.

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 $$0$$.

• 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. – ZeroTheHero Oct 27 '20 at 12:25
• @ZeroTheHero $\langle n |m \rangle$ is a number so $\dagger$ and $*$ coincide. – jacob1729 Oct 27 '20 at 13:27