# Annihilation and Creation operators not hermitian

The annihilation and creation operators are given below $$\hat a|n\rangle=\sqrt{n}|n-1\rangle\qquad\text{and}\qquad\hat a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$$

Now I know the operator a^ is not Hermitian but as to why I'm a little confused. Is it because they 'create' and 'annihilate' photons hence they're not hermitian? I don't think that explanation would be good for an exam if this question was asked though...

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Is this in the context of QM or QFT? Anyway, you could prove it from the definition of the operators. Other options would be because if they were hemitian, $a=a^\dagger$, which wouldn't make sense . Or since the Hamiltonian is a sum of positive terms, the only way to factor it must be using complex operators. –  jinawee Jun 8 '14 at 8:34
This is in terms of quantum mechanics and if I was to consider it from the properties of the a operators –  Kennan Jun 8 '14 at 8:41
And are the states normalized and can you use the commutation relations? –  jinawee Jun 8 '14 at 9:02
Well if assuming the only information is what has been given in my post (what the operators are) and assumed knowledge of what the defining properties of a^ and a^* are –  Kennan Jun 8 '14 at 9:12

Creation and annihilation operators are ladder operator in the sense that they raise and lower respectively the quantum numbers of a state (such as e.g. the number of particles in an harmonic oscilattor, the angular momentum for spins, etc...). If they were hermitian, that is $a=a^\dagger$, the same operator $a$ should lower and raise the quantum number at the same time spoiling its very definition. Perhaps, you can think of the number'' operator $a^\dagger a$ as a sort of hermitian cousin of $a$ or $a^\dagger$, and indeed it does not raise or lower the quantum numbers but it simply counts.

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If the dagger in $a^\dagger$ were just a label this wouldn't be a proof. –  jinawee Jun 8 '14 at 9:25
@jinawee good point: just to confirm: you are saying that whereas this answer does prove that the raising and lowering operators cannot be the same, it does not prove that $a^\dagger$ is Hermitian, right? –  WetSavannaAnimal aka Rod Vance Jun 8 '14 at 9:57
@WetSavannaAnimalakaRodVance I'm confused by your wording. This answer assumes $a^\dagger$ is the hermitian conjugate of $a$, which implies that the hermitian property is meaningless (except if $\vert n\rangle=0$). But it doesn't prove it if rasing and lowering operators are not the hermitian conjugate of each other (I think that's what you meant). Otherwise, the excercise would be trivial. In fact, the OP didn't say if $\vert n\rangle$ is a vector, so I think the (rigorous) proof doesn't exist. –  jinawee Jun 8 '14 at 10:06
@jinawee What I mean is the argument does show something: that $a$ and $a^\dagger$ cannot be the same (or, if they were, they wouldn't do what we would intuitively want them too). But, as you say, if one isn't going to beg the question, you must look at $\dagger$ as just a label, or, better still as in your answer, write $b = a^\dagger$, then what the answer shows is clear, and it's simply that $b\neq a$. –  WetSavannaAnimal aka Rod Vance Jun 8 '14 at 10:21
@jinawee Why should I supposed that $(a^\dagger)^\dagger$ is not $a$? To me, by definition, annihilation/creation operators are a pair $(a,b)$ where $b\equiv a^\dagger$ satisfying $[a,a^\dagger]=1$. Not sure what definition you are instead assuming. Anyway, the OP asked whether it can happen than the pair is actually made of hermitian operators, that is $a^\dagger=a$. I have explained by words why it isn't the case. But of course from the algebra $[a,a^\dagger]=1$ it is even more apparent that $a$ can't be equal to $a^\dagger$. –  TwoBs Jun 8 '14 at 12:34

We have the annihilation and creation operators $a$ and $a^\dagger$, respectively (we know that they are the hermitian conjugates of each other, but we won't assume that fact). So let's rename $a^\dagger=b$.

I'll try to show that $a=a^\dagger$ and $b=b^\dagger$, gives a contradiction. We know:

$$ba\vert n\rangle=n\vert n\rangle \quad, \quad ab\vert n\rangle=(n+1)\vert n\rangle$$We are supposing that $a^\dagger$ is not the hermitian conjugate of $a$.

With this you could prove: $[a,b]=\mathbb{I}$

Then:

$$n^2\langle n \vert n\rangle=\langle n\vert a^\dagger b^\dagger ba\vert n \rangle=\langle n\vert a b ba\vert n \rangle=$$

$$=\langle n\vert[( b^\dagger a^\dagger+\mathbb{I}) ba]\vert n \rangle=n(n+1) \langle n\vert n \rangle+n \langle n\vert n \rangle$$

Where I have used the commutation relation and the hermiticy property. This yields: $n=0$ or $\vert n\rangle =0$.

So unless $\vert n\rangle=\vert 0\rangle$ or $0$, $a$ and $a^\dagger$ won't be hermitian (I don't think it holds if the only state is $\vert 0 \rangle$ or $0$ but I'm not sure).

Is it because they 'create' and 'annihilate' photons

This reasoning would be very sloppy. Ladder operators arise in the context of the harmonic oscillator, angular momentum... But in Quantum Mechanics, particle number is conserved (you could say something like it modifies the energy, hence it emmits a photon or something like that). These operators create and destroy photons in Quantum Field Theory.

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+1, this is correct whilst the other answer decidely is not, for the reason ion your comment. I suggest a change your first sentence: "That $a\neq a^\dagger$ is proven by contradiction as follows, therefore, let's assume to contrary and rename $a^\dagger=b$, where ...": otherwise I get thoroughly confused by the first sentence. –  WetSavannaAnimal aka Rod Vance Jun 8 '14 at 10:21

My first reaction is to give an argument along the lines of the other answers, if $a=a^\dagger$ then saying that $a$ raises and $a^\dagger$ lowers doesn't make sense.

However, if you don't want to assume that the hermitian conjugate of $a$ is a raising operator, you could be more direct than that though and show $a$ is not hermitian given just that $a$ is a lowering operator, without knowing $a^\dagger$ is a raising operator. All you have to do is construct the matrix elements of $a$ in some convenient basis, and show that the resulting matrix is not hermitian. The $|n\rangle$ basis is a pretty convenient one!

In order for $a$ to be hermitian, it must be that $a_{mn}^*= a_{nm}$, or equivalently $$\langle m | a | n \rangle^* {=}^? \langle n | a| m \rangle$$ But this is clearly false given the definition of $a$, because $$LHS = \langle m | a | n \rangle^* = \sqrt{n} \langle m | n-1 \rangle^* = \sqrt{n} \delta_{m,n-1} = \sqrt{n} \delta_{m+1,n}$$ while $$RHS = \langle n | a | m \rangle = \sqrt{m} \langle n-1 | m \rangle = \sqrt{n-1} \delta_{m-1,n}$$

Clearly $LHS \neq RHS$, so $a$ is not hermitian. [To see they don't equal just try a few examples, any time $LHS\neq 0$ then $RHS=0$. If you wrote them out in matrix notation, LHS would have entries one slot above the diagonal, RHS would have entries one slot below.]

In fact, this is essentially a proof that $a^\dagger$ is a raising operator (since $\langle n | a^\dagger | m \rangle = \langle m | a | n \rangle^*$), so this argument is not fundamentally that different from the other answers.

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This is my favorite argument, as it stems directly from the definition of the operator and the definition of hermiticity. –  David Z Jun 8 '14 at 23:28

Lets give an answer based on the physical meaning of the operator $\hat{a}$

The operator $\hat{a}$, is the quantum-mechanical operator which represents a measure of the classical complex representation of the electric field. $\hat{a}$ in classical physics is well-known as the analytic signal, which is in short a generalization of the complex representation of any time-varying signal.

For example: $$2*E_0cos(\omega t) = E_0e^{i\omega t} + E_0e^{-i \omega t}$$ The complex representation of the real harmonic signal $E_0 cos(\omega t)$ can be taken as $E_0/2e^{i\omega t}$. But what to pick as the complex representation for any time-varying electric field ? Well, one can see from the simple above example that the complex representation can be defined by keeping only the positive frequency part of the Fourier transform of the signal.

This complex number (the analytic signal a) for any real signal x(t), can be written as $a = x - ip$ where $p$ deduces from the Hilbert transform of $x(t)$

Quantum mechanics is simply the mathematical reinterpretation of the cinematics of the classical physics quantities and by construction the eigenvalues of the operator $\hat{a}$ are complex.

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the eigenvalues of the operator $a^\dagger$ are complex No, they are real. measure of the classical complex representation of the electric field In the harmonic oscillator there is no electric field in principle, so I don't know what you mean. –  jinawee Jun 9 '14 at 22:31
I brought the essence of the physical meaning of the annihilation operator in the context of the question. How do you get the two quadratures of the oscillator, if the values were real ? Do you assume some sort of representation ? The eigenstates of the annihilation operator are more than well-known to be the coherent states. en.wikipedia.org/wiki/Coherent_states Read the father of modern quantum optics if you want to understand the meaning of the annihilation operator and see that my answer is correct. journals.aps.org/pr/abstract/10.1103/PhysRev.130.2529 –  JumpArtist Jun 10 '14 at 9:35
$\hat a|n\rangle=\sqrt{n}|n-1\rangle$ The operators are complex, but $n$ is a real (integer) number, so the eigenvalues are real. –  jinawee Jun 10 '14 at 11:52
$|n\rangle$ is not an eigenstate of the annihilation operator, but an eigenstate of the number operator $\hat{N}$. The annihilation operator $\hat{a}$ does not commute with $\hat{N} = \hat{a}^\dagger \hat{a}$. One has $\hat{a}|\alpha\rangle = \alpha |\alpha\rangle$ where $\alpha$ is a complex number and the eigenvalue associated to the eigenstate $|\alpha\rangle$ of the operator $\hat{a}$. $|\alpha\rangle$ is well-known to be decomposed as an infinite superposition of Fock states $|n\rangle$. One would see then that the mean value of $\hat{N}$ in an eigenstate of $\hat{a}$ is $|\alpha|^2$. –  JumpArtist Jun 12 '14 at 19:37
To add details, $|\alpha|^2$ would correspond to the mean value of a Poisson distribution of Fock states. The eigenstates of $\hat{a}$ do not have well-defined photon number (see the above non-commutation). Again, please read en.wikipedia.org/wiki/Coherent_states –  JumpArtist Jun 12 '14 at 20:09