# Momentum operator and eigenvalue

Can we take the eigenvalue we obtained after momentum operator (of quantum mechanics) operates on the state vector of a system as the momentum of the system?

• In the spirit of this: "the system can always be said to 'have' a value of momentum" is false. – Emilio Pisanty Nov 11 '19 at 15:48
• The best you can do in dreaming about the momentum of a state is looking at the expectation value of the momentum operator in that state. – Cosmas Zachos Nov 11 '19 at 16:17
• Thank you so much – SHYAMANANDA NINGOMBAM Nov 12 '19 at 10:37

No. Mathematically multiplying your state vector by an operator is not how you determine the outcome of a measurement. First, that would assume you can get a deterministic value, which we know isn't true of measurements of quantum systems. Second, if your system is (most likely) in a superposition of momentum states, then this operation will not even give you a single value for momentum, as your state is not an eigenstate of momentum.

The only way this works is if your state was an eigenstate of momentum before the momentum measurement. Then by coincidence operating on your state vector with the momentum operator will give you the momentum of the system as that eigenvalue. However, states like these are not physical, so this case cannot be realized.

• Thank you so much... Then.. After operating momentum operator on a system ,which is not its eigen state, we got a value which is similar to eigen value. Than what does this value give/signify? – SHYAMANANDA NINGOMBAM Nov 12 '19 at 10:36
• @SHYAMANANDANINGOMBAM If it's not an eigenstate then you will not get a value that is like an eigenvalue. – Aaron Stevens Nov 12 '19 at 14:53
• Okay... But... If we get some value from the calculation... Does it have any physical significance? – SHYAMANANDA NINGOMBAM Nov 12 '19 at 15:18
• @SHYAMANANDANINGOMBAM You wouldn't get any value. You would get some other vector. But you could, for example, use that to get the expectation value. i.e. $\langle|\psi|P|\psi\rangle$ – Aaron Stevens Nov 12 '19 at 15:49

Yes, if it's in a pure momentum eigenstate, which is the only way you would obtain a single eigenvalue of the momentum operator. There are complications due to the fact that you can't really have a system in a pure momentum state (though you can get arbitrarily close). But conceptually, that's what it means to have a single eigenvalue.

Note: applying the momentum operator does not change the state of the system. The way to do that, i.e., get a system into a pure eigenstate, is to perform a measurement, which collapses the wavefunction.

• Thank you so much... Then.. After operating momentum operator on a system ,which is not its eigen state, we got a value which is similar to eigen value. Than what does this value give/signify? – SHYAMANANDA NINGOMBAM Nov 12 '19 at 10:39
• The set of eigenstates form a basis. You'll get a of superposition of eigenstates, the coefficients of which form the probability amplitude. – Richter65 Nov 12 '19 at 12:16