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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 the Hilbert space $\mathcal{H}$ in question?

For a while I thought that the answer is yes, but if I think about a finite-dimensional Hilbert space I can never apply a matrix (a.k.a, some representation of $a_{j}^{\dagger}$) to the zero vector, and get out a vector that is not the zero vector. However, you can do this with the vacuum state.

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marked as duplicate by ACuriousMind, Danu, Kyle Kanos, JamalS, Qmechanic Mar 27 '15 at 23:14

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    $\begingroup$ $\left|0\right>$ is not necessarily the zero vector. I suppose you could have a situation in which the two are equal, but not in the general case $\endgroup$ – Jim Mar 27 '15 at 16:45
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    $\begingroup$ no, the zero vector does not correspond to any physical state, not even vacuum/ground state. $\endgroup$ – image Mar 27 '15 at 16:45
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    $\begingroup$ It cannot be the zero vector otherwise $a^\dagger_j |0> = |0>$ just by linearity! $\endgroup$ – Valter Moretti Mar 27 '15 at 17:09
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Don't be fooled by the zero inside the ket. That is just a label. For example in Scalar QFT, the vacuum state of interacting theories is usually denoted as $|\Omega\rangle$ rather than $|0\rangle$. So no, the vacuum state does not represent the null vector of the Hilbert space.

Rather, the vacuum state is defined to be the state with the lowest possible energy, so that an application of the annihilation operator gives zero, i.e,

$$ \hat{a} |0\rangle = 0$$

so that this state has the lowest possible non-zero energy,

$$ \hat{H}|0\rangle = \frac{1}{2}\hbar\omega|0\rangle $$

Edit: The zero vector is just a mathematical object with no physical interpretation. The zero vector is formally defined as the additive identity of the additive group so that,

$$\mathbf{u} + \mathbf{0} = \mathbf{u}$$

for all vectors $\mathbf{u}$.

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No. I might be misremembering my linear algebra here, but the zero vector should be additive identity, correct? But the state $| 0 \rangle + | 1 \rangle$ is clearly distinct from the state $| 1 \rangle$, as can be shown by taking expectation values for any observable that differs between vacuum and first excited state.

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Your concern is justified; a linear operator will always have an output of zero when the input is a zero vector. However, when you operate on the vacuum state with the annihilation operator, you do get a zero vector. What the vacuum state look like will then depend on your representation of creation and annihilation operators, but the rule that $a_j|0>=0$ will give you the answer.

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The vacuum state is (or mathematically looks like) the ground state of a ficticious harmonic oscillator. So you have a state you might label $| 0 \rangle$, but that is not a zero (0), but a basis state, the one for zero quanta in the harmonic oscillator.

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