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$(1)$Since quantum-mechanical states between two consecutive measurements are represented as superposition of orthonormal basis vectors in a vector space, at first glance it seems like it makes sense to talk about a state like $ \frac{1}{√2}|{\uparrow}{\rangle}+\frac{1}{√2}|{\downarrow}{\rangle}$ as something completely distinct from either $|{\uparrow}{\rangle}$ or $|{\downarrow}{\rangle}$ (which form the orthonormal basis), sort of like any $3$-vector can be written as a sum of unit basis vectors multiplied by scalars and yet is distinct from any of them.

$(2)$However, you can often find a description of superposited states as a system being in both basis states at the same time (e.g. the spin being simultaneously up and down), but I have never found a similar description regarding vector quantities (for example velocity being along $z$ axis and $x$ axis at the same time if the vector was somewhere between these axes) in classical mechanics, despite both situations being a result of addition of the basis vectors.

Why? Is the problem more complicated than what I deduced in $(1)$?

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    $\begingroup$ I'm not sure what you mean by "classical superposition". In classical mechanics, you cannot add states because the states do not, in general, form a vector space. So there is no classical analogue to being in a superposition of two states. $\endgroup$ – ACuriousMind Oct 20 '15 at 21:02
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    $\begingroup$ I mean a superposition of vectors. Not specifically state-vectors, all vectors. $\endgroup$ – Qwedfsf Oct 20 '15 at 21:05
  • $\begingroup$ What's the status of this question? Do you need more information? If so, please explain what is left to understand. $\endgroup$ – DanielSank Oct 22 '15 at 4:52
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This is all just a result of sloppy language on the part of people describing quantum mechanics. The state $$ \left\lvert \Psi \right\rangle = \frac{1}{\sqrt{2}} \left( \left\lvert \uparrow \right\rangle + \left\lvert \downarrow \right\rangle\right) \tag{1}$$ is a superposition of the two orthogonal states $\left\lvert \uparrow \right\rangle$ and $\left\lvert \downarrow \right\rangle$. The state is unlike either basis vector alone. A velocity vector $$\left\lvert v \right\rangle = a\left\lvert x \right\rangle + b\left\lvert y \right\rangle \tag{2}$$ for some values $a$ and $b$ is also a superposition of two orthogonal velocity vectors. It is unlike either basis vector alone.

Talking about $\left\lvert \Psi \right\rangle$ as "simultaneously in both states" is just plain sloppy. It's a superposition. It's not like either basis vector alone. It is, as you say, something completely distinct.

However, you can often find a description of superposited states as a system being in both basis states at the same time (e.g. the spin being simultaneously up and down), but I have never found a similar description regarding vector quantities (for example velocity being along z axis and x axis at the same time if the vector was somewhere between these axes) in classical mechanics, despite both situations being a result of addition of the basis vectors.

The reason for this disagreement in language comes from the fact that, in the end, quantum state vectors tell you probabilities of experimental outcomes. It really bugs people to think of the state of a physical system being fundamentally probabilistic. When it comes to measurement, the state $\left\lvert \Psi \right\rangle$ means that the system has a 1/2 probability to be measured spin up and and 1/2 probability to be measured spin down. People don't naturally think about the world around them in terms superposition states whose coefficients correspond to probability amplitudes. They'd rather think about the classical states independently and try to form some kind of notion of the system existing in combinations of classical states. Therefore, they naturally (but erroneously) say that the system is in both classical states at the same time, when really, as you said, the system is in a state that's completely different from either classical basis state.

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  • $\begingroup$ Hi, sir; as you said the actual state is different from the base states that contribute in the superposition; may I ask you how they are different? For me they are different as they have different expected values of the observables, isn't it? $\endgroup$ – user36790 Oct 21 '15 at 10:02
  • $\begingroup$ @user36790 This is unfortunately a very broad question. You are right that the expectation values of various measurable quantities are different, but there are more differences as well. $\endgroup$ – DanielSank Oct 21 '15 at 15:29
  • $\begingroup$ Sorry, sir, if it's a broad query. However, could you please mention one apart from the expectation issue? I'm self-studying QM. So, if you help, I would be grateful: just one:) $\endgroup$ – user36790 Oct 21 '15 at 15:31
  • $\begingroup$ @user36790 Why not ask a question as a new post? $\endgroup$ – DanielSank Oct 21 '15 at 15:49
  • $\begingroup$ Okay, then; I would do it; thanks for the advice, sir:) $\endgroup$ – user36790 Oct 21 '15 at 15:54
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One of the more bizarre features of quantum mechanics is quantum superposition of states, which is distinctly different from classical superposition of conventional vectors. Classical superposition requires linear summations of vectors such as position vectors, or sinusoids represented as vectors. In classical superposition, if you add a position vector to another position vector you get a third position vector that represents the net position of an object. But quantum superposition is superposition of states (also usually represented as vectors), whereby a coin can actually be in a state that is a summation of both heads and tails, or a cat can be both alive and dead (before measurement, i.e., observation). No one knows what this superposition of states really means, but it is the only way to explain certain quantum phenomena (at least so far), such as quantum entanglement. There is no classical analog to superposition of states. And BTW, Richard Feynman is still right, no one fully understands quantum mechanics.

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I encountered the similar question. I think one should distinguish the difference between classical mixture (i.e. bunch of particle $A$ and particle $B$ put together) and quantum superposition. The difference there was significant, i.e. for classical mixture $N=N_1+N_2$, for quantum superposition $|\psi_i\rangle =\frac{1}{\sqrt{2}}|A\rangle+\frac{1}{\sqrt{2}}|B\rangle $.

The difference between quantum superposition and classical superposition, however, did not have much difference except the physical interpretation of orthogonal states, i.e. particle go through slit 1 and slit 2 at the same time. Because essentially quantum superposition was developed based on classical superposition, and until now the mathematical description was just a form of linear operator.

The real difference of quantum superposition arise from the causality, i.e. it proved that the indefiness of the causality, or say non causal model. [A comparative terms was the counterfactual indefineness and non locality proved by quantum entanglement.] It required us to rewrite the model and reference frame. However, those concepts has only been recently proved and become popular after 2016. Therefore as far as the textbook concerned, just put some fancy words and wild imagination from thought experiments.

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