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Thoughts about ∣Φ>=∣1>*C1+∣2>*C2

  1. Aug 14, 2007 #1
    Hi, we know that in a system with two base states, any state ∣Φ> of the system can always be described as a linear combination of the two base state:∣Φ>=∣1>*C1+∣2>*C2, where C1 is the ampitudes to be in on base state and C2 the other.
    My question is, does that mean the state ∣Φ> can only swing (flip) between the two base state ∣1> and∣2>? I mean, is there any other actually new state except the two base states? If not, then why is it that we can choose two other new base state ∣Ⅰ>and∣Ⅱ>and rewrite the ∣Φ> in the new form :∣Φ>=∣Ⅰ>*CⅠ+∣Ⅱ>*CⅡ ?
    Thank you in advance :)
  2. jcsd
  3. Aug 15, 2007 #2
    if you have studied the linear algebra, you should know the rank of this space is 2
  4. Aug 15, 2007 #3
    Yeah, I studied that and I think a n-state system in QM can be analogous to a n-rank space, right?

    But my question is not this, Let me put my it this way: since ∣Φ>=∣1>*C1+∣2>*C2, which means that the state ∣Φ> has the amplitude C1 to be in the base state ∣1> and the amplitude C2 to be in the base state ∣2>.Then what’s the state ∣Φ> exactly? As we know, any vector in a coordinate can be analyzed into two base unit vector (e.g. e1 for x-direction and e2 for y-direction) and in fact this vector does indeed exist and stay the same no matter what coordinate we choose to represent it. I know in QM any state ∣Φ> is analogy to a “vector”, but the amplitude doesn’t sound very comfortable(we can’t analyze the state, but just talk about the amplitudes and they are not exactly the same as projection). I mean, is the state ∣Φ> actually indeed a definite state that exists and stays unchanged like a vector in coordinate? What exact is it? 
  5. Aug 15, 2007 #4

    why cant we analyze the state,we can analyze this state in the position representation or monentum representation or so.in my opinion,the state,which involves all the information of a system,exists really in a system.and if you want to know the position distribution of the system,you can use the position representation to reach it,or you can get the momentum distribution using the momentum representation.maybe i still dont understand your literally mean,but i indeed think the state does indeed just like the vector and exist really.
  6. Aug 15, 2007 #5
    In QM, we only measure a state using the definition of amplitude, not simply analyze it like what we do in classical physics.
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