Quantum Computing and Superposition of states

  • #1
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Main Question or Discussion Point

I'm watching a lecture on the intro to quantum computing.

See the attached image which will be useful as I describe my question.

So the professor says that we have this single photon and it's in this state, ## | 0 > ##.

He states that when we send this photon through a beam splitter that it will (assuming we're not doing any sort of measurement), end up in a superposition state of

[tex] \frac{1}{\sqrt{2}}( |0> + |1>) [/tex]

If you look at beam splitter 1 (BS1) in the image, you'll see the incoming photon in state ## |0> ## and two outgoing states, ## |0> ## to the right and ## |1>## up.

My question is this: considering this superposition state, does it even make sense to think about these two different paths? Or is EACH outgoing path through BS1 this same superposition state of

[tex] \frac{1}{\sqrt{2}}( |0> + |1>) [/tex]

Put another way, should I basically think about the beamsplitter as this black box that takes an input and just outputs a single superposition state into the next beam splitter?

Thanks a lot.
 

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  • #2
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I'm watching a lecture on the intro to quantum computing.

[]

My question is this: considering this superposition state, does it even make sense to think about these two different paths? Or is EACH outgoing path through BS1 this same superposition state of

[tex] \frac{1}{\sqrt{2}}( |0> + |1>) [/tex]

Put another way, should I basically think about the beamsplitter as this black box that takes an input and just outputs a single superposition state into the next beam splitter?

Thanks a lot.
A quantum beam splitter has two input channels and two output channels. The diagram looks wrong. One of the inputs is being assumed and you need to know what it is.

See the 'Quantum Description' section here https://en.wikipedia.org/wiki/Beam_splitter#Quantum_mechanical_description
 
  • #3
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It may help here if you think of a photon as of a classical wave. Then a superposition of the states is just a sum of two waves - one going right and another going up. So the superposition applies to the "global" state, considering both paths, but if you want to consider somehow the state along one path, it is not a superposition (in this particular experiment), but either |0> or |1>.
 
  • #4
138
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It may help here if you think of a photon as of a classical wave. Then a superposition of the states is just a sum of two waves - one going right and another going up. So the superposition applies to the "global" state, considering both paths, but if you want to consider somehow the state along one path, it is not a superposition (in this particular experiment), but either |0> or |1>.
Thanks @MichPod. I still have some confusion about these outbound paths that he labled |0> and |1>. Those paths are only really defined if we actually MAKE a measurement correct? That is, if we setup a detector that tries to detect which outbound path the photon is coming from, THEN we will have these paths as |0> and |1>. Is that correct?

And then second question: in the case where we don't try and detect the path, then as you said it's this superposition. But my confusion there lies in the fact that there are still TWO inputs into the second beam splitter. What do those two inputs represent?

Thanks again for the help.
 
  • #5
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These "paths" are |0> and |1> by themselves, unrelated to whether any measurement is done or not. Again, taking a classical pucture, you can say that you have some wave propagating right and some wave propagating up. Or you can talk of them as of one wave propagating both up and right (a superposition).

The actual state of the photon is a superposition, but if you put a mirror on some path, that mirror will affect only one part of the superposition - the same way as a mirror put on the way of the classical wave will affect only the part of the wave which collided with a mirror.

Before you even talk about "detection" which is a quantum phenomena, try to fully understand how a classical light would propagate through the system. This system is known as https://en.m.wikipedia.org/wiki/Mach–Zehnder_interferometer and must be described in many places on internet, including some very basic layman level.
 
  • #6
Strilanc
Science Advisor
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does it even make sense to think about these two different paths? Or is EACH outgoing path through BS1 this same superposition state
You will have a very hard time if you try to think of each outgoing path's state on its own.

Basically, in order to do so, you'll have to understand mixed states, partial traces, and second quantization. You'll have to start thinking of the superpositions as this:

##\frac{1}{\sqrt{2}}|\text{left}_{\text{photon=yes}}\rangle |\text{right}_{\text{photon=no}}\rangle + \frac{1}{\sqrt{2}}|\text{left}_{\text{photon=no}}\rangle |\text{right}_{\text{photon=yes}}\rangle##

then, in order to focus on one path, you'll have to "trace out" the other path. This only works if you convert to a density matrix first. If you trace out ##|\text{right}_*\rangle## you get:

##0.5 |\text{left}_{\text{photon=no}}\rangle\langle\text{left}_{\text{photon=no}}| + 0.5 |\text{left}_{\text{photon=yes}}\rangle\langle\text{left}_{\text{photon=yes}}|##

Which is equivalent to a classical probability distribution. Not because the state is actually random instead of under superposition, but because the process we used throws away the information necessary to understand in what way the state is quantum instead of probabilistic.
 

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