# Can “state” be considered a 5th dimension?

I searched for an answer to this question on Google but just found articles that mention either string theory or a 5th dimension in passing (such as Maxwell equations as they relate to Riemann curvature tensor.)

I stumped myself while driving to school today thinking about this... We can explain an objects position in the universe by describing its spacial and time locations, such as at 2nd and 3rd street on the fifth floor at 10:00am

But is that enough to fully describe an object when you take into account Schrödinger's cat? For example, what if, Schrödinger's cat is the object that is at that location and time. It seems like you would have to have a 5th state dimension to fully explain its position, as in at 2nd and 3rd street on the fifth floor at 10:00am with a probability of .5

It seems like if we had state as an additional dimension it would help explain things like quantum entanglement as the particles could be moving away from each other in space-time but standing still in state.

Am I merely relating things that have no business being related, or is there a connection between state and dimension?

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In your given example, Schr\"{o}dinger's Cat, the sate is discrete, "dead" or "alive". It would not be a continuous dimension. –  Flint72 Apr 18 '14 at 22:20
Have you heard about phase space? –  Anixx Apr 18 '14 at 22:24
@Flint72 So the cat does not exist in both states at once? I was under the impression that it did, which led me to my question in the first place. –  ialexander Apr 19 '14 at 0:15
I hadn't heard of phase space. Reading up on it now... –  ialexander Apr 19 '14 at 0:16

It goes the other way, actually. In the Lagrangian and Hamiltonian approaches to classical dynamics (on which the quantum theory is based), you learn about "generalized coordinates" or "degrees of freedom." Most often this is used to reduce the complexity of a problem. The canonical example is a clock pendulum, constrained to move in a plane. The motion in $x$ and $y$ is rather complicated, but it becomes simpler once you realize that the only degree of freedom is the angle $\theta$ that the pendulum makes with the vertical direction.