Examining Coherent States: a|0⟩=0|0⟩=0

[LagrangeEuler]

Coherent states are eigen state of lowering operator $a$
|\alpha\rangle=e ^{-\frac{|\alpha|^2}{2}}\sum^{\infty}_{n=0}\frac{\alpha^n}{\sqrt{n!}}|n \rangle, where $\{|n \rangle\}$ are eigenstates of energy operator. What is the case of state $|0 \rangle$?
a|0 \rangle=0|0 \rangle=0.
So, $|0 \rangle$ is eigenstate of lowering operator. But how to get that from
|\alpha\rangle=e ^{-\frac{|\alpha|^2}{2}}\sum^{\infty}_{n=0}\frac{\alpha^n}{\sqrt{n!}}|n \rangle ?

 

[stevendaryl]

Well, $|0\rangle = lim_{\alpha \rightarrow 0} |\alpha\rangle$

 

[Cthugha]

What is the case of state $|0 \rangle$?
a|0 \rangle=0|0 \rangle=0.
So, $|0 \rangle$ is eigenstate of lowering operator.

Stop! This is incorrect.
a|0 \rangle=0 is correct

a|0 \rangle=0|0 \rangle is explicitly not the case! You cannot subtract a photon (I assume we are talking about the light field here) from the vacuum. You discard this state completely.

The closest physical implementation to the photon annihilation operator would be a beam splitter with a very small reflection ratio placed inside a light beam and a detector in the arm of the reflected light that clicks whenever it gets hit by a photon. If you now investigate the light field in the transmission arm conditioned on a detection event in the reflection arm, you will get a good approximation to a state which corresponds to the photon annihilation operator acting on the initial state. However, if the initial state is the vacuum, that detector will never ever click...

 

[hutchphd]

Stop! This is incorrect.

a|0⟩=0a|0⟩=0​

a|0 \rangle=0 is correct

a|0⟩=0|0⟩​

Pardon my ignorance but then how does one apply the number operator a^†a to the ground state ?? Your formalism seems screwy.

​

 

[Cryo]

Pardon my ignorance but then how does one apply the number operator a†a to the ground state ?? Your formalism seems screwy.

​

One doesn't, or more precisely if one does one gets zero. For systems such as harmonic oscillator, the Hamiltonian is $H\propto \hat{a}^\dagger \hat{a}+\frac{1}{2}$, so you get non-zero energy even for ground-state. For systems with continuum of states, such as light in vacuum, you get a divergence of the ground-state energy. I am sure there are plenty of people here who can explain how to live with this divergence better than me, but as far as I remember, you simply igonre this divergence by talking about energies relative to ground-state energy. So then, by definition the energy of the ground state is zero.

 

[Cthugha]

Pardon my ignorance but then how does one apply the number operator a^†a to the ground state ?? Your formalism seems screwy.

​

Well, you directly get the 0 out. However, if this seems too strange for you, you can still use the fact that both operators are bosonic, so that a^†a= a a^†-1. If you evaluate the operators this way, you get:
n=\langle 0|a a^\dagger |0 \rangle-1= \langle 1| 1|1\rangle-1=0

Sure, the formalism may look strange at first, but it is a direct consequence of the fact that the annihilation operator is not like a typical operator representing a measurement. You can always, say, perform a position measurement, so you can always apply the position operator, but you cannot always employ the annihilation operator in a physically meaningful way. If there is nothing to annihilate, there is just no way to perform annihilation of a particle in a meaningful way.

 

[hutchphd]

Thank you for the insight. Never really worried about it before...very interesting

 

[stevendaryl]

The technical meaning of an "eigenstate" of an operator $A$ is a state $|\psi\rangle$ such that there is a real or complex number $\lambda$ such that $A |\psi\rangle = \lambda |\psi\rangle$. I've never heard anyone say that $\lambda$ cannot be 0.

 

[Cthugha]

Well, yes, this may seem kind of strange, but essentially this is the same physics you see for ladder operators for the harmonic oscillator. They connect two states of adjacent occupation number and as the occupation -1 just does not exist, you also just get 0 when you apply the ladder operator to the ground state and not 0 times the ground state.

This is more or less a natural consequence of taking a discrete variable (which means: photon numbers) approach towards describing coherent states. If you consider continuous variables (which means field quadratures) instead, it is much easier to see the connection between the vacuum state and a coherent state, which is just a displaced vacuum. That can be seen nicely in the Wigner or Husimi-Q functions of those fields. All the reasonable experiments that investigate the vacuum in optics were also performed using continuous variable. The field operator is a sum of the creation and annihilation operator as opposed to a product in discrete variables, so it can always be implemented in experiment in contrast to discrete variables, where photon annihilation from the vacuum just cannot be realized in a closed system in free space.

 

[stevendaryl]

Well, yes, this may seem kind of strange, but essentially this is the same physics you see for ladder operators for the harmonic oscillator. They connect two states of adjacent occupation number and as the occupation -1 just does not exist, you also just get 0 when you apply the ladder operator to the ground state and not 0 times the ground state.

I don't understand that last sentence. According to the rules of multiplication, $0 \cdot |0\rangle = 0$. So I don't understand the distinction you're making.

 

[stevendaryl]

I don't understand that last sentence. According to the rules of multiplication, $0 \cdot |0\rangle = 0$. So I don't understand the distinction you're making.

Oh, what is different for the ground state is that the rule for lowering operators doesn't hold:

$a |n\rangle = \sqrt{n} |n-1\rangle$

Since there is no state $|-1\rangle$, this doesn't hold when $n=0$ (unless you have some kind of multiplication law that says $0 \cdot undefined = 0$.)

 

[Cthugha]

Yes, indeed. That is the problem. And: sorry. I now noticed that my notation was pretty unclear. Sorry for that. I would like to emphasize that the 0 you get is not the complex number 0, but rather the null vector of the Hilbert space, which is a huge difference.

 

[vanhees71]

I think this is again a nice example for a misunderstanding due to notation ;-). The vacuum state, let's call it $|\Omega \rangle$ to avoid the trouble the entire thread suffers from, is of course and eigenstate of $\hat{N}=\hat{a}^{\dagger} \hat{a}$ with eigenvalue zero.

An operator acting on a vector gives a vector. Now acting with $\hat{N}$ (or $\hat{a}$ for that matter) you get also a vector, the null vector of the Hilbert space, which I'd denote with $|0 \rangle$, but now I'm in a dilemma, if I denote the eigenstates of $\hat{N}$ with $|n \rangle$, because now $n \in \{0,1,2,\ldots \}$. Thus it's more clever to denote the eigenvectors with $|u_n \rangle$. Then you have $|u_0 \rangle=|\Omega \rangle$ and $\hat{a} | u_0 \rangle=|0 \rangle$ :-).

 

[hutchphd]

I think this is again a nice example for a misunderstanding due to notation ;-). The vacuum state, let's call it $|\Omega \rangle$ to avoid the trouble the entire thread suffers from, is of course and eigenstate of $\hat{N}=\hat{a}^{\dagger} \hat{a}$ with eigenvalue zero.

An operator acting on a vector gives a vector. Now acting with $\hat{N}$ (or $\hat{a}$ for that matter) you get also a vector, the null vector of the Hilbert space, which I'd denote with $|0 \rangle$, but now I'm in a dilemma, if I denote the eigenstates of $\hat{N}$ with $|n \rangle$, because now $n \in \{0,1,2,\ldots \}$. Thus it's more clever to denote the eigenvectors with $|u_n \rangle$. Then you have $|u_0 \rangle=|\Omega \rangle$ and $\hat{a} | u_0 \rangle=|0 \rangle$ :-).

So is the null vector included in every basis of the space? Is it normalized?

 

[vanhees71]

The null vector exists in any vector space. The most trivial vector space consists only of the null vector. By definition it's the vector space of dimension 0.

The null vector can never be included in a basis since by definition a basis is a complete set of linearly independent vectors, i.e., if $|a_k \rangle$ is a basis, then if
$$\sum_k \lambda_k |a_k \rangle =0$$
one must have $\lambda_k=0$ for all $k$. Thus $|0 \rangle$ cannot be in the basis.

 

[hutchphd]

The null vector exists in any vector space. The most trivial vector space consists only of the null vector. By definition it's the vector space of dimension 0.

The null vector can never be included in a basis since by definition a basis is a complete set of linearly independent vectors, i.e., if $|a_k \rangle$ is a basis, then if
$$\sum_k \lambda_k |a_k \rangle =0$$
one must have $\lambda_k=0$ for all $k$. Thus $|0 \rangle$ cannot be in the basis.

Should this not read
$$\sum_k \lambda_k |a_k \rangle = |0 \rangle$$

I may be caught in a semantic trap here but I thought a basis spanned the space. I recall this means (by definition) that every element can be expressed in terms of the basis. Is this not the definition?

 

[vanhees71]

It should indeed read
$$\sum_k \lambda_k |a_k \rangle = |0 \rangle$$
where $|0 \rangle$ is the null vector.

Concerning the basis, it's even better: Every element of the vector space can be expressed uniquely as a linear combination of the basis vectors. That's the definition of "basis". Put differently: A basis is a complete set of linearly independent vectors.

 

[hutchphd]

It should indeed read
$$\sum_k \lambda_k |a_k \rangle = |0 \rangle$$
where $|0 \rangle$ is the null vector.

Concerning the basis, it's even better: Every element of the vector space can be expressed uniquely as a linear combination of the basis vectors. That's the definition of "basis". Put differently: A basis is a complete set of linearly independent vectors.

But then how do we represent $|0 \rangle$ as such a linear combination? Surely not that sum with all λ_k = 0 ? Hence my original question

So is the null vector included in every basis of the space?

 

[Dr.AbeNikIanEdL]

Surely not that sum with all λk = 0 ?

Why not?

 

[vanhees71]

Indeed, as I said the decomposition of any vector as the linear combination of basis vectors by definition is unique, and this means that the one and only decomposition of the null vector is to set all its components with respect to any basis to 0. That's the very definition of a set of linearly independent vectors. Thus a basis is by definition a complete set of linearly independent vectors, where complete means that any vector can be written as a linear combination of this set.

In short: A basis is any complete set of linearly independent vectors.

 

[hutchphd]

But as you pointed out above this is not the vacuum state, right? Does the null state provide any physical manifestation? Does any operation on the null state produce nonzero result? Thanks.

 

[stevendaryl]

But as you pointed out above this is not the vacuum state, right? Does the null state provide any physical manifestation? Does any operation on the null state produce nonzero result? Thanks.

The state of a system has to be nonnull. The null state is there in the algebra of states (any linear combination of states is another state), but it can't be the actual state of a physical system.

 

[LagrangeEuler]

Stop! This is incorrect.
a|0 \rangle=0 is correct

a|0 \rangle=0|0 \rangle is explicitly not the case! You cannot subtract a photon (I assume we are talking about the light field here) from the vacuum. You discard this state completely.

The closest physical implementation to the photon annihilation operator would be a beam splitter with a very small reflection ratio placed inside a light beam and a detector in the arm of the reflected light that clicks whenever it gets hit by a photon. If you now investigate the light field in the transmission arm conditioned on a detection event in the reflection arm, you will get a good approximation to a state which corresponds to the photon annihilation operator acting on the initial state. However, if the initial state is the vacuum, that detector will never ever click...

Yes, but coherent state are also states in which $\Delta x \cdot \Delta p=\frac{1}{2}\hbar$. And it is true for state $|0\rangle$. Right? Light is example. But you can have coherent states also in condensed matter...

 

[LagrangeEuler]

We can also put question in different form. Find eigenvalues and normalized eigenvectors of operator
b=\frac{1}{\sqrt{2}}(\xi-\frac{d}{d \xi})
where $\xi=\sqrt{\alpha}x=\sqrt{\frac{m \omega}{\hbar}}x$. So is $\psi_0(x)=(\frac{\alpha}{\pi})^{\frac{1}{4}}e^{-\frac{\alpha x^2}{2}}$ eigenvector of $b$?

 

[Cthugha]

Yes, but coherent state are also states in which $\Delta x \cdot \Delta p=\frac{1}{2}\hbar$. And it is true for state $|0\rangle$. Right? Light is example. But you can have coherent states also in condensed matter...

Sure, but this is a different property. Your question was whether applying the photon annihilation operator leaves the vacuum intact. This is not the case. Of course you can also define the set of coherent states as the set of all displaced vacuum states, which you get by applying the displacement operator to the vacuum, where the displacement will correspond to the coherent amplitude.

We can also put question in different form. Find eigenvalues and normalized eigenvectors of operator
b=\frac{1}{\sqrt{2}}(\xi-\frac{d}{d \xi})
where $\xi=\sqrt{\alpha}x=\sqrt{\frac{m \omega}{\hbar}}x$. So is $\psi_0(x)=(\frac{\alpha}{\pi})^{\frac{1}{4}}e^{-\frac{\alpha x^2}{2}}$ eigenvector of $b$?

This is somewhat strange notation. Are you sure that there is no typo in there? Essentially, you seem to write the annihilation operator and the vacuum/coherent state in the coordinate representation. This is discussed e.g. in chapter 2.3. ("what is so coherent about coherent states?") of the quantum optics book by Scully and Zubairy. In a nutshell: applying the annihilation operator in this notation to the vacuum state gives you a constant 0 as the result (which is also the reason why it is often denoted as the number zero, although I find this notation misleading).

 

[vanhees71]

Of course, the ground state is an eigenstate of $\hat{b}$ with eigenvalue 0. Just calculate the derivative to see it.

 

[Cthugha]

Of course, the ground state is an eigenstate of $\hat{b}$ with eigenvalue 0. Just calculate the derivative to see it.

Now I am somewhat puzzled. Are you sure about that? Maybe I just missed something, but isn't $\hat{b}$ the equivalent of the creation operator? I always thought that the annihilation operator had the plus sign.

 

[vanhees71]

You are right. I've misread the sign.

The annihilation operator is
$$\hat{a}=\sqrt{\frac{m \omega}{2}} \hat{x}+\mathrm{i} \frac{1}{\sqrt{2m \omega}} \hat{p}.$$
In position representation that maps to
$$\hat{a}=\sqrt{\frac{m \omega}{2}} x +\frac{1}{\sqrt{2 m \omega}} \partial_x = \frac{1}{\sqrt{2}} (\xi + \partial_{\xi}).$$
The creation operator is its adjoint,
$$\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\xi-\partial_{\xi})=\hat{b},$$
which of course has no eigenstate.

The annihilation operator has many eigenstates, the coherent states. The special one with eigenvalue 0 is the ground state. This gives the eigenvector equation
$$\hat{a} u_0(\xi)=0 \; \Rightarrow \; \partial_{\xi} u_0=-\xi u_0 \xi \; \Rightarrow \; u_0(\xi)=N \exp(-\xi^2/2)$$
with $N$ a normalization factor.