Quantum States and ladder operator

[kashokjayaram]

In any textbooks I have seen, vacuum states are defined as:
a |0>= 0
What is the difference between |0> and 0?

Again, what happens when a⁺ act on |0> and 0?

and Number Operator a⁺a act on |0> and 0?

 

[mfb]

0 is a real number, like 1 and $\pi$
Edit: Sorry not here, see post #5.

|0> is a state. The digit inside is just a convenient name. You could name it |myfavoritestate> and nothing would change (well, the formulas would look more complicated).

Operators "act on" states. If you write a⁺ 0 (or any other operator instead of a⁺), this is not an operator acting on a state, it is an operator multiplied with a real number - and a multiplication with 0 has the result 0.

Again, what happens when a+ act on |0>

You get another state, called |1>.

and Number Operator a+a act on |0>

You get the number of the state |0>, which is 0 (hence the name of the state).

 

[stevendaryl]

Really, you should first thoroughly understand the way raising and lowering operators work in the non-relativistic quantum mechanics of the Harmonic Oscillator. There, you start with observables $x$ and $p$, and you make the switch to ladder operators

$A = \sqrt{\frac{m \omega}{2 \hbar}} x + i \sqrt{\frac{1}{2 m \omega \hbar}} p$
$A^\dagger = \sqrt{\frac{m \omega}{2 \hbar}} x - i \sqrt{\frac{1}{2 m \omega \hbar}} p$
In terms of $A$ and $A^\dagger$, the corresponding states are:

$|n\rangle$ where

$A\ |n \rangle = \sqrt{n}\ |n-1\rangle$
$A^\dagger\ |n\rangle = \sqrt{n+1}\ |n+1\rangle$

The state $|0\rangle$, when you translate back into $x$ language, is something like $C e^{-\lambda x^2}$ for appropriate constants $\lambda$ and $C$. It's not zero, it's just that the operator $A$ acting on it produces 0.

 

[dauto]

|0> represents a physical state. 0 is just a number.

 

[Bill_K]

0 is a real number, like 1 and $\pi$

|0> represents a physical state. 0 is just a number.

Not in this case. When you act on a state like |0> with an operator like a, you do not get a number, you get another element of the Hilbert space.

In the example given, a |0>= 0, the 0 on the right hand side is the zero element of the Hilbert space.

 

[mfb]

Ah, right. Sorry.

 

[kashokjayaram]

In the example given, a |0>= 0, the 0 on the right hand side is the zero element of the Hilbert space.

Zero element of the Hilbert space means what??
|0> is also the element of Hilbert space...! Isn't it?
Is it the "smallest element(norm of which is small)"..??
Or why is then a acts gives zero..??

Sorry I can't distinguish... that's why..!

 

[vanhees71]

No, $|0 \rangle$ is the vacuum (ground) state. It's not the null vector of the Hilbert space. That's why often one better writes another symbol, e.g., $|\Omega \rangle$ for the vacuum or ground state, which is a state of minimal energy.

 

[WannabeNewton]

Or why is then a acts gives zero..??

Because $|0\rangle$ is the lowest possible energy eigenstate of $\hat{H}$. You can't go any lower. $\hat{a}$ is a lowering operator that takes you from excited states to less excited states but if $|0 \rangle$ is the lowest possible mode excitation of the harmonic oscillator then $\hat{a}$ can't take you any lower than that so it must annihilate $|0 \rangle$ i.e. $\hat{a} |0 \rangle = 0$.

But $|0\rangle$ is not the zero vector. In fact if you solve for $\psi_0(x) = \langle x | 0 \rangle$ in the differential equation $\langle x | \hat{a} | 0 \rangle = x\psi_0 + x_0^2 \frac{d\psi_0}{dx} = 0$ you'll get a Gaussian.

 

[stevendaryl]

Zero element of the Hilbert space means what??
|0> is also the element of Hilbert space...! Isn't it?
Is it the "smallest element(norm of which is small)"..??
Or why is then a acts gives zero..??

Sorry I can't distinguish... that's why..!

If $|\psi\rangle$ is any state, and $\alpha$ is any number (real or complex), then there is another state $|\psi'\rangle = \alpha |\psi\rangle$. The "zero element" 0 is what you get when you choose $\alpha = 0$.

The state $|0\rangle$ is not the zero element. It's a state with no particles in it--the vacuum. The operator $N = a^\dagger\ a$ is the "number operator". The state $|n\rangle$ is the eigenstate with $n$ particles. It satisfies the equation:

$N\ |n\rangle = n\ |n \rangle$

So the state $|0\rangle$ is the state satisfying $N |0\rangle = 0 |0\rangle =$0.

 

[kashokjayaram]

If $|\psi\rangle$ is any state, and $\alpha$ is any number (real or complex), then there is another state $|\psi'\rangle = \alpha |\psi\rangle$. The "zero element" 0 is what you get when you choose $\alpha = 0$.

The state $|0\rangle$ is not the zero element. It's a state with no particles in it--the vacuum. The operator $N = a^\dagger\ a$ is the "number operator". The state $|n\rangle$ is the eigenstate with $n$ particles. It satisfies the equation:

$N\ |n\rangle = n\ |n \rangle$

So the state $|0\rangle$ is the state satisfying $N |0\rangle = 0 |0\rangle =$0.

Why is it another state...?? $\alpha$ is a number which doesnot change the vectors in vector space. So it will be the same state. Am I right..?
What happens when $a^\dagger$ act on 0??

 

[stevendaryl]

Why is it another state...?? $\alpha$ is a number which doesnot change the vectors in vector space. So it will be the same state. Am I right..?
What happens when $a^\dagger$ act on 0??

Well, there is an ambiguity about what "the same state" means. A lot of quantum mechanics uses normalized states, where you multiply by a constant so that $\langle \Psi | \Psi \rangle = 1$. But the zero vector cannot be normalized. It's not equal to any other state.

And operating on the zero vector with any operator just returns the zero vector again.

 

[vanhees71]

In standard quantum theory the pure states are equivalently described by either rays in Hilbert space or the special class of statistical operators that are projection operators.

A ray in Hilbert space is given by an equivalence class of non-zero vectors. Two non-zero vectors [itex]|\psi_1 \rangle and $\psi_2 \rangle$ belong to the same class if and only if there is a non-zero complex number $\lambda$ such that $|\psi_2 \rangle=\lambda |\psi_1 \rangle$. Sometimes (and again equivalently) for convenience one restricts oneself to normalized state vectors only. Then two state vectors are in the same ray if they differ by a phase factor only, i.e., a complex number of modulus 1.

A Statistical operator is any self-adjoint positive semidefinite operator $\hat{R}$ with $\mathrm{Tr} \hat{R}=1$. It represents a pure state if and only if it is a projection operator, i.e.,
if additionally $\hat{R}^2=\hat{R}$.

It's clear that both definitions of a pure state are equivalent. For each representant $|\psi \rangle \neq 0$ the operator
$$\hat{R}=\frac{1}{\|\psi \|^2} |\psi \rangle \langle \psi|$$
fulfills the requirements of a Statistical operator, representing a pure state.

If, on the other hand, $\hat{R}$ represents a pure state in the above given sense, then due to $\hat{R}^2=\hat{R}$ it has only eigenvalues $0$ and $1$. Since at the same time $\mathrm{Tr} \hat{R}=1$ there is one and only one eigenvector with eigenvalue $1$. Choosing it normalized $\langle \psi|\psi \rangle$ this implies that the statistical operator must be the projection operator
$$\hat{R}=|\psi \rangle \langle \psi|.$$

It is important to keep in mind that a pure state is represented not exactly by a (normalized) Hilbert-space vector but by an entire ray (normalized Hilbert-space vector modulo an arbitrary phase factor), because it implies important subtle points: e.g., it admits that also half-integer eigenvalues make sense for angular momenta and thus that particles with half-integer spins exist. Among others these are the electrons, protons, and neutrons, making up all everyday matter!

 

[stevendaryl]

It is important to keep in mind that a pure state is represented not exactly by a (normalized) Hilbert-space vector but by an entire ray (normalized Hilbert-space vector modulo an arbitrary phase factor), because it implies important subtle points: e.g., it admits that also half-integer eigenvalues make sense for angular momenta and thus that particles with half-integer spins exist. Among others these are the electrons, protons, and neutrons, making up all everyday matter!

That's a leap that's too big for me to follow. Why does the use of rays, rather than vectors, imply anything about half-integer spins? Oh, maybe because you don't require that a 360 degree rotation is the identity, but only that it brings you another vector in the same ray?