Linked Questions

3
votes
4answers
2k views

Understanding the Quantum Vacuum State [duplicate]

In terms of the creation and annihilation operators $a_{j}$ and $a_{j}^{\dagger}$ (fermionic or bosonic, doesn't matter): Is the vacuum state $\mid\mathrm{vacuum}\rangle$ exactly the zero vector on ...
17
votes
1answer
5k views

After using annihilation operator on vacuum state, why it is $0$ instead of vacuum?

For bosonic systems, why $a|0\rangle=0$ and not $a|0\rangle=|0\rangle$?
9
votes
3answers
2k views

Why particle number operator $\hat{N}$ is $\hat{a}^\dagger\hat{a}$ rather than $\hat{a}\hat{a}^\dagger$?

Both $\hat{a}^\dagger\hat{a}$ and $\hat{a}\hat{a}^\dagger$ are Hermitian, how do we know which one represents the particle number?
6
votes
4answers
589 views

Difference between kets $\left| -x\right\rangle$ and $-\left |x\right\rangle$?

While using dirac bra-ket notation for quantum mechanics, what's the difference if the minus sign is inside or outside the ket ? I know that $\left| -x\right\rangle$ and $- \left|x \right\rangle$ ...
2
votes
3answers
425 views

What is the physical meaning of $a_{\vec{p}} \! \mid \! 0 \rangle$

$a^\dagger_{\vec{p}} \! \mid \! 0 \rangle = \mid \! p \rangle$ is interpreted as a creation of a particle with momentum $p$ from the vacuum. $a_{\vec{p}} \! \mid \! p \rangle = \mid \! 0 \rangle$ is ...
8
votes
3answers
728 views

Is $0 | \psi \rangle=0$?

For example, the spin operator for spin 1 particle is $$ \hat{S}_z\doteq\hbar\begin{pmatrix} 1&&\\&0&\\&&-1\end{pmatrix} {\qquad} \text{for state} {\qquad} \left|\psi\right>...
1
vote
2answers
639 views

Inner product in QFT

When we write inner product in QM for example $\langle\psi\rvert x \psi\rangle$it means (in position space) $\int\psi^*(x,t)x\psi(x,t)dx$. But when we write, in QFT, $\langle0\rvert0\rangle=1$ what ...
0
votes
1answer
145 views

Difference between $|0\rangle$ and $0$ in the context of isospin

I know this has been asked before, but I am confused having read it in the context of isospin, where the creation operators act on the "vacuum" state (representing no particles) $$a^\dagger_m|0\rangle ...
1
vote
1answer
166 views

Ground state Dirac formalism

In Dirac formalism, the action of the 'lowering' operator acting on a ground state energy eigenstate is given by $$\hat{a}|0\rangle=0$$ Notation wise, it is clear what the left hand side express ...
2
votes
1answer
155 views

What's the equation $\hat{a}|0\rangle=0$ saying?

This is a well-known result of ladder operators, which obviously means that you can't remove energy from the vacuum. But what is $\hat{a}|0\rangle=0$ actually saying? How does it say the "you can't" ...