1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Thoughts about ∣Φ>=∣1>*C1+∣2>*C2
  1. Spin 1/2 correlation (Replies: 6)

  2. Spin 2 1/2 (Replies: 4)

  3. Spin 1/2 (Replies: 12)