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I am a beginner in Quantum mechanics and as I understand,the superposition of stationary states is also a solution of time-independent Schrödinger equation (TISE). The wave functions that are the solutions of TISE are stationary states. So this should mean the superposition of stationary state is also a stationary state. But it turns out it is not as the probability density varies with time if the energies of each state are different. I am terribly stuck after reading a lot of books so here's my question:

Is the superposition of stationary states a stationary state? If no, how can the solution to a TISE not be a stationary state (as I see in all the physical problems we find solutions to TISE and they are stationary)?

Also, I read somewhere that the expected value of Hamiltonian stays constant with time for such superposed wavefunction. If the expectation of Hamiltonian is constant then why isn't the superposed state stationary?

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    $\begingroup$ Superposition of stationary states is not a stationary state(correct me if I'm wrong); if measurement is done, then only the superposed state collapses to one of the stationary states; before that it was in superposition. $\endgroup$ – user36790 Oct 6 '15 at 17:21
  • $\begingroup$ In short, TISE solutions $|E_1⟩$ and $|E_2⟩$ with different energies $E_1≠E_2$ are solutions of different equations,$$\hat H|E_1⟩=E_1|E_1⟩\ \text{and}\ \hat H|E_2⟩=E_2|E_2⟩.$$ Superposing solutions of the same linear equation gives you a solution of that equation. Superposing solutions of different equations doesn't need to give anything special. $\endgroup$ – Emilio Pisanty Oct 6 '15 at 19:05
  • $\begingroup$ Which books you have read? look at example 2.1 Griffiths 4th edition and the ans is NO. $\endgroup$ – Paul Oct 8 '15 at 3:56
  • $\begingroup$ The answer is not so correct. Consider an eigenspace ${\cal H}_E$ of the Hamiltonian with energy $E$, but with dimension $n>1$. Let $\psi$ and $\psi'$ two linearly independent normalized vectors in ${\cal H}_E$. Both are stationary states and any (normalized) linear combination of them is still a stationary state with the same energy $E$. A linear combinations of stationary states is not a stationary state if the energies of the states are different. $\endgroup$ – Valter Moretti Oct 8 '15 at 8:04
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A superposition of stationary states is not a stationary state. Suppose we have two kets, $| E_1 \rangle$ and $|E_2 \rangle$, which solve the TISE like so: $$ \hat{H} | E_i \rangle = E_i | E_i \rangle $$ Here, the left hand side is the Hamiltonian operator and and the right hand side just shows that the stationary state picks up an eigenvalue. This is the definition of a solution of the TISE, although you may be more familiar with the differential operator version which is equivalent to left-multiplying by a position eigenstate $\langle x |$. Now, suppose we create a new superposition state, $| \psi \rangle$, defined as: $$ | \psi \rangle = \frac{1}{\sqrt{2}} \left( | E_1 \rangle + | E_2 \rangle \right) $$ If this is a solution to the TISE (that is, if it is a stationary state/eigenstate of the Hamiltonian) then it should follow that $\hat{H} | \psi \rangle = c | \psi \rangle$, where $c$ is a constant. But we can use linearity to write: $$ H | \psi \rangle = \frac{1}{\sqrt{2}} \left( \hat{H} | E_1 \rangle + \hat{H} | E_2 \rangle \right) \\ = \frac{1}{\sqrt{2}} \left( E_1 | E_1 \rangle + E_2 | E_2 \rangle \right) $$ This is not a constant $c$ multiplied by the vector $| \psi \rangle$ unless $E_1 = E_2$. So we can see that two stationary states cannot be combined into another stationary state unless they share the same eigenvalue.

As for the expectation value of the Hamiltonian, let's evaluate that real quick: $$ \langle \hat{H} \rangle = \langle \psi | \hat{H} | \psi \rangle \\ = \frac{1}{2} \left( \langle E_1 | + \langle E_2 | \right) \hat{H} \left( | E_1 \rangle + | E_2 \rangle \right) \\ = \frac{1}{2} \left( \langle E_1 | \hat{H} | E_1 \rangle + \langle E_2 | \hat{H} | E_1 \rangle + \langle E_2 | \hat{H} | E_2 \rangle + \langle E_1 | \hat{H} | E_2 \rangle \right)\\ = \frac{1}{2} \left(E_1 + E_2 \right) $$ Nothing about time dependence here at all. Basically, a combination of energy eigenkets will have a Hamiltonian expectation value which is a weighted average of the energies. Since the relative weights of energy eigenkets won't change, it stays constant--and we can always decompose a ket into a sum of energy eigenkets, so this is always true. (Assuming a time-independent Hamiltonian.) Note that this doesn't apply for other operators necessarily, since if it doesn't share eigenkets with the Hamiltonian you might find that $\langle E_1 | \hat{A} | E_2 \rangle$ does not evaluate to zero for an arbitrary operator $A$.

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  • $\begingroup$ You missed $|$ in $|E_2\rangle$; BTW, +1. $\endgroup$ – user36790 Oct 6 '15 at 18:44
  • $\begingroup$ Fixed. Sometimes I actually put a line of \newcommand{\ket} into my answers, and sometimes I just wing it... $\endgroup$ – zeldredge Oct 6 '15 at 18:48
  • $\begingroup$ Yes, I noticed it in one of your answers to my question...the answer was great, BTW:) $\endgroup$ – user36790 Oct 6 '15 at 18:54
  • $\begingroup$ Unless the stationary states are degenerate (share the same energy level), in which case the superposition is going to be again a stationary state? (Which is contained in the answer, but deserves to be highlighted, IMHO). $\endgroup$ – Frank Dec 2 '17 at 18:03
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It is true that any linear combination of solutions of a linear equation is still a solution to the same linear equation. However, when you are talking about the TISE, the equation itself can be different object if the Hamiltonian is changed. In that case, the superposition of two solution states of two different TISEs may not be a solution of another TISE with the some given Hamiltonian. The superposition states of stationary is still stationary only if you are using the same Hamiltonian for all the scenarios you are talking about.

Is this answering your question? Or, you may want to make your question a little bit clear.

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  • $\begingroup$ So does this mean, TISE doesn't always have solutions as stationary states? $\endgroup$ – Dodo Oct 6 '15 at 17:39
  • $\begingroup$ A TISE always has stationary solutions by definition. It is just the way how you find those stationary solutions is a question in my opinion. $\endgroup$ – X Qi Oct 6 '15 at 17:50
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A stationary state is described as follows. I take the example of the energy observable. The total energy operator , i.e. the Hamiltonian has n eigen values, say, each corresponding to a different stationary state. If the system is in one of the stationary states, then you can perform a huge number of experiments on the system to measure the energy of the system, you will always get the same measurement, i.e. the eigen value of the corresponding stationary state eigen function.

Now, when you consider the superposed state of the n corresponding eigen functions, created by a linear combination of them, you no more have a single eigen value corresponding to a single eigen function, instead you have n eigen values of the energy observable. So now if you conduct a large number of experiments on the system, you will get different values at different times (provided you make the measurements sufficiently spaced in time), each with a different probability proportional to the coefficient of the corresponding eigen function in the linearly superposed mixed state. Hence now you'll have a specific expectation value which is expected to be different from the different individual eigen values. But that doesn't mean you are going to get the same value of energy for each and every iteration of the experiment. You are actually going to get different energy values, whose 'average' would be the expectation value. So the superposed state is not a stationary state. This would answer your first question as well. Since the energy observed on measuring a system in a superposed state is not a fixed 'stationary' value, this eigen function would not be a solution of the TISE.

Hope this helps.

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  • $\begingroup$ So as time advances, the expectation(the average) of all of the values remains constant? Beacuse the sum of all of the energy eigenvalues of all the including states will be constant. Is it so? $\endgroup$ – Dodo Oct 6 '15 at 17:55
  • $\begingroup$ The eigen values of the eigen states are all constant over time, as a result of which the expectation value comes out to be a constant. I reiterate that expectation value is just a probabilistic value. $\endgroup$ – Arkya Chatterjee Oct 6 '15 at 18:08

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