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$\newcommand{\k}[1]{\left | #1 \right\rangle }$ Dirac in his book The Principles of Quantum mechanics states that:

To proceed with the mathematical formulation of the superposition principle we must introduce a further assumption, namely the assumption that by superposing a state with itself we cannot form any new state, but only the original state over again. If the original state corresponds to the ket vector $\k{ A} $, when it is superposed with itself the resulting state will correspond to $$c_1 \k A + c_2\k A = (c_1 + c_2) \k A $$ where $c_1$ and $c_2$ are complex numbers. Now we may have $c_1 + c_2 = 0$, in which case the result of the superposition process would be nothing at all, the two components having cancelled each other by an interference effect. Our new assumption requires that, apart from this special case, the resulting state must be the same as the original one, so that $(c_1 + c_2)\k A $ must correspond to the same state that $\k A $ does.

And then again he quips in the last few lines of the same paragraph and the next paragraph:

Thus a state is specified by the direction of a ket vector and any length one may assign to the ket vector is irrelevant. All the states of the dynamical system are in one - one correspondence with all the possible directions for a ket vector, no distinction being made between the directions of the ket vectors $\k A $ and $- \k A .$ The assumption just made shows up very clearly the fundamental difference between the superposition of the quantum theory and any kind of classical superposition. In the case of a classical system for which a superposition principle holds, for instance a vibrating membrane, when one superposes a state with itself the result is a different state, with a different magnitude of the oscillations. There is no physical characteristic of a quantum state corresponding to the magnitude of the classical oscillations, as distinct from their quality, described by the ratios of the amplitudes at different points of the membrane. Again, while there exists a classical state with zero amplitude of oscillation everywhere, namely the state of rest, there does not exist any corresponding state for a quantum system, the zero ket vector corresponding to no state at all.

My questions are:

  1. In the first paragraph, if $c_1 + c_2 = 0$ in $c_1\k A + c_2\k A = (c_1 + c_2) \k A $, then there should be a destructive interference as is said. But this also implies \begin{align}c_1\k A + c_2\k A &= c_1\k A - c_1\k A \\ &= c_1(\k A -\k A )\\ &= c_1(\k A + (-\k A ))\\ &=0\end{align} So for $(\k A + (-\k A )) = 0$, $\k A $ and $-\k A $ must have opposite sense in some way at least. But the magnitude of $\k A $ is irrelevant as mentioned and the directions of $\k A $ and $-\k A $ are not distinct.

Then how are $\k A $ and $-\k A $ different notations? And how do they have opposite sense to satisfy the above relation?

  1. Why does the zero ket vector correspond to no state? The last 2 lines of the paragraph look very confusing indeed.
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  • $\begingroup$ Please try not to use newcommand, as that forces everyone on this page to use your commands (could be worse commands than you give). You can just use \rangle without issue. $\endgroup$ – Kyle Kanos Dec 29 '15 at 20:16
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First of all, I'd like to convey what Dirac was trying to convey, in my own words. What he is talking about in the first paragraph is the principle of normalization. Mathematics is a tool for describing physics. Just because Schrödinger's equation is satisfied by physical systems doesn't mean that all solutions of the equation are physical. We need to normalize the solutions, i.e., the solutions should have a magnitude of 1 (this is related to Born's probability rule).

Now if you notice, all vectors $c_1 \k A$ and $c_2 \k A$, when normalized, will become the same thing. For instance, assume that the magnitude of $\k A$ is 1. Then the factors $c_1$ and $c_2$ have to 1. If they aren't, then they aren't physical and we neglect them.

Now, answering your questions:

  1. If you read paragraph 1 carefully, Dirac says "apart from this special case", which means that we have to exclude the case of $c_1 + c_2 = 0$. So your first problem turns into a no problem. Now, the $\k A$ and $(-\k A)$ vectors are the same because when we normalize them, the negative sign goes away. We write $$-\k A = e^{i \pi} \k A$$ and this phase factor is unmeasurable in physical situations.

  2. The zero ket is by definition a ket of zero magnitude. This ket is 'non-normalizable', i.e., it's magnitude cannot be made 1. It's probability of occurrence is 0. Hence, it is a non-state, since it doesn't exist (0 probability = doesn't exist).

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  1. A global complex coefficient in front of a vector is not relevant for defining the corresponding state. However the relative coefficients in a superposition are crucial for the definition of the state. As a matter of fact, writing a suitable superposition of vectors it is possible to obtain any vector of the Hilbert space, and therefore any (pure) state.

  2. The zero vector $\lvert 0\rangle$ is such that $\langle 0,0\rangle=\lVert 0\rVert^2=0$, and that does not agree with the probabilistic interpretation of a state, since it must have total probability (i.e. norm) one.

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  • $\begingroup$ Can you explain what you mean by global and relative coefficients? I just started studying quantum mechanics by reading this book. I have very elementary knowledge. So can you ease it a bit so I can understand? $\endgroup$ – SchrodingersCat Dec 28 '15 at 11:29
  • $\begingroup$ A global coefficient is a coefficient that multiplies the whole vector, as the $z\in\mathbb{C}$ in $z(\lvert a_1\rangle + \dotsc + \lvert a_n\rangle)$. The relative coefficients are the $(c_n)_{n\in\mathbb{N}}\subset \mathbb{C}$ that appear in the superposition $\sum_{n=0}^{N}c_n \lvert a_n\rangle$ (where $N\leq \infty$). $\endgroup$ – yuggib Dec 28 '15 at 11:33
  • $\begingroup$ Okay. And shouldn't $|A>$ and $-|A>$ be opposite in sense as I mentioned? $\endgroup$ – SchrodingersCat Dec 28 '15 at 11:38
  • $\begingroup$ I think that the common intuition about vectors should be taken with due care when dealing with (infinite dimensional) Hilbert spaces and quantum mechanics. The "ket" vectors are a useful tool to represent states, but it has to be kept in mind that the states are a probabilistic object that describes how observables would be measured in the system. And it is not so meaningful to define the "sense" or "direction" of a probability... $\endgroup$ – yuggib Dec 28 '15 at 11:48
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A state is a normalised positive linear functional over the algebra of observables. Given a representation $\pi$ of such algebra, all vector states are of the form $$(v,\pi(A)v)$$ for a unit vector $v$ from the representation Hilbert space $H_\pi$. The set of all the intertwiners $(\pi,\pi)$ can be identified with the commutant of the representation. In particular, every partial isometry $V\in(\pi,\pi)$ that contains the vector $v$ in its initial domain is such that $$(Vv,\pi(A)Vv) = (v,\pi(A)v).$$ Hence the unit vectors associated to vector states are unique up to partial isometries. If the representation $\pi$ is irreducible, as in the case of the Schroedinger representation, the algebra of observables is usually assumed to be the whole of $B(\ell^2(\mathbb N))$, and $(\pi,\pi)$ reduces to the multiples of the identity operator. In this case, vectors corresponding to states are defined up to a scalar factor in $\mathbb C$, usually called a phase. Hence $v$ and $-v$ define the same state, as well as $zv$ for any $z\in\mathbb C$ such that $|z|=1$.

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